NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions
1. SEO Strategy Introduction & Chapter Master Overview
Chapter 5, Arithmetic Progressions (AP), forms one of the most foundational, high-scoring algebraic pillars in the CBSE Class 10 Mathematics curriculum. In the standard CBSE Board Examination blueprint, questions from Arithmetic Progressions consistently account for 4 to 6 marks, spanning Multiple Choice Questions (1 mark), Short Answer Questions (2 or 3 marks), and high-weightage Case Study-based Questions (4 marks). Mastery over Arithmetic Progressions establishes the mathematical framework required for understanding higher-level sequences, series, discrete mathematics, and financial algorithms in senior secondary classes.
The conceptual core of an Arithmetic Progression lies in the concept of a constant rate of discrete change. An Arithmetic Progression is defined as a sequence of numbers in which each term is obtained by adding a fixed number, termed the common difference ($d$), to the preceding term, except the first term ($a$). A comprehensive analysis of past decade CBSE papers reveals that examiners design questions targeting three primary areas: identifying valid AP sequences from practical scenarios, determining unknown parameters via the $n^{\text{th}}$ term formula ($a_n = a + (n – 1)d$), and computing series totals using the summation formula ($S_n = \frac{n}{2}[2a + (n – 1)d]$).
A frequent pitfall among Class 10 students is confusing the term index $n$ with the term value $a_n$. Students often forget that the number of terms $n$ must strictly be a positive integer ($n \in \mathbb{N}$); fractional or negative values of $n$ render a physical term non-existent within the sequence. Another common misconception occurs during the calculation of terms from the end of an AP, where students incorrectly apply the forward common difference rather than reversing the sequence or adjusting the sign of $d$.
To secure maximum marks under official CBSE marking criteria, answers must present a structured deductive workflow: explicitly state the Given Data, identify and state the Standard Mathematical Formula, execute Step-by-Step Algebraic Simplification, and conclude with an Unambiguous Final Statement with appropriate units. High-scoring students systematically verify boundary conditions, such as rejecting negative roots in quadratic equations when solving for the count of terms $n$.
Plaintext
+----------------------------------------------------------------------------------------------------+
| MASTER SUMMARY: ARITHMETIC PROGRESSIONS |
+-----------------------------------+-----------------------------------+----------------------------+
| Concept / Parameter | Mathematical Formula / Condition | Key Scoring Notes |
+-----------------------------------+-----------------------------------+----------------------------+
| General Form of an AP | a, a + d, a + 2d, a + 3d, ... | 'a' = first term |
| | | 'd' = common difference |
+-----------------------------------+-----------------------------------+----------------------------+
| Common Difference (d) | d = a_{k+1} - a_k | Can be positive, negative, |
| | | or zero |
+-----------------------------------+-----------------------------------+----------------------------+
| n-th Term (General Term) | a_n = a + (n - 1)d | n must be a natural |
| | | number (n in N) |
+-----------------------------------+-----------------------------------+----------------------------+
| n-th Term from the End (Last = l) | a_n' = l - (n - 1)d | 'l' is the last term |
+-----------------------------------+-----------------------------------+----------------------------+
| Sum of First n Terms (Standard) | S_n = (n / 2) * [2a + (n - 1)d] | Used when 'd' is known |
+-----------------------------------+-----------------------------------+----------------------------+
| Sum of First n Terms (Last Term) | S_n = (n / 2) * (a + l) | Used when first and last |
| | | terms are known |
+-----------------------------------+-----------------------------------+----------------------------+
| Relation Between a_n and S_n | a_n = S_n - S_{n-1} | Valid for all n >= 2; |
| | | a_1 = S_1 |
+-----------------------------------+-----------------------------------+----------------------------+
| Three Consecutive Terms in an AP | (a - d), a, (a + d) | Common difference is 'd' |
+-----------------------------------+-----------------------------------+----------------------------+
| Four Consecutive Terms in an AP | (a - 3d), (a - d), (a + d), | Common difference is '2d' |
| | (a + 3d) | |
+-----------------------------------+-----------------------------------+----------------------------+
| Arithmetic Mean (AM) of a and c | b = (a + c) / 2 | If a, b, c are in AP, |
| | | then 2b = a + c |
+-----------------------------------+-----------------------------------+----------------------------+
2. Exercise 5.1 Solutions (Page No. 99-100)
Question 1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes $\frac{1}{4}$ of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.
(iv) The amount of money in the account every year, when ₹ 10,000 is deposited at compound interest at $8\%$ per annum. (Page No. 99)
Answer:
(i) Let $a_n$ denote the total taxi fare for travelling a distance of $n\text{ km}$.
According to the given condition:
$$\begin{aligned} a_1 &= \text{Fare for } 1\text{ km} = \text{₹ } 15 \\ a_2 &= \text{Fare for } 2\text{ km} = 15 + 8 = \text{₹ } 23 \\ a_3 &= \text{Fare for } 3\text{ km} = 23 + 8 = \text{₹ } 31 \\ a_4 &= \text{Fare for } 4\text{ km} = 31 + 8 = \text{₹ } 39 \end{aligned}$$
Now, calculating the successive differences:
$$\begin{aligned} a_2 – a_1 &= 23 – 15 = 8 \\ a_3 – a_2 &= 31 – 23 = 8 \\ a_4 – a_3 &= 39 – 31 = 8 \end{aligned}$$
Since $a_{k+1} – a_k = 8$ is constant for all values of $k$, the list of numbers $15, 23, 31, 39, \dots$ forms an Arithmetic Progression (AP) with first term $a = 15$ and common difference $d = 8$.
(ii) Let the initial volume of air present in the cylinder be $V$ units.
In each stroke, the vacuum pump expels $\frac{1}{4}$ of the air remaining in the cylinder.
$$\begin{aligned} a_1 &= V \\ a_2 &= V – \frac{1}{4}V = \frac{3}{4}V \\ a_3 &= \frac{3}{4}V – \frac{1}{4}\left(\frac{3}{4}V\right) = \frac{3}{4}V\left(1 – \frac{1}{4}\right) = \left(\frac{3}{4}\right)^2 V = \frac{9}{16}V \\ a_4 &= \frac{9}{16}V – \frac{1}{4}\left(\frac{9}{16}V\right) = \left(\frac{3}{4}\right)^3 V = \frac{27}{64}V \end{aligned}$$
Checking the differences between consecutive terms:
$$\begin{aligned} a_2 – a_1 &= \frac{3}{4}V – V = -\frac{1}{4}V \\ a_3 – a_2 &= \frac{9}{16}V – \frac{3}{4}V = -\frac{3}{16}V \end{aligned}$$
Since $a_2 – a_1 \neq a_3 – a_2$, the difference between successive terms is not constant. Hence, the given situation does not form an AP (it forms a Geometric Progression).
(iii) Let $a_n$ represent the total cost of digging the well up to a depth of $n$ metres.
Given:
$$\begin{aligned} a_1 &= \text{Cost for digging the } 1^{\text{st}}\text{ metre} = \text{₹ } 150 \\ a_2 &= \text{Cost for digging } 2\text{ metres} = 150 + 50 = \text{₹ } 200 \\ a_3 &= \text{Cost for digging } 3\text{ metres} = 200 + 50 = \text{₹ } 250 \\ a_4 &= \text{Cost for digging } 4\text{ metres} = 250 + 50 = \text{₹ } 300 \end{aligned}$$
Evaluating the common differences:
$$\begin{aligned} a_2 – a_1 &= 200 – 150 = 50 \\ a_3 – a_2 &= 250 – 200 = 50 \\ a_4 – a_3 &= 300 – 250 = 50 \end{aligned}$$
Since the consecutive difference $a_{k+1} – a_k = 50$ is constant throughout, the sequence $150, 200, 250, 300, \dots$ forms an AP with $a = 150$ and $d = 50$.
(iv) Let $P = \text{₹ } 10000$ be the principal amount, and $r = 8\%$ per annum be the rate of compound interest.
The total amount $A_n$ in the account at the end of year $n$ is given by the compound interest formula $A_n = P\left(1 + \frac{r}{100}\right)^n$.
$$\begin{aligned} a_1 &= 10000\left(1 + \frac{8}{100}\right)^1 = 10000(1.08) = 10800 \\ a_2 &= 10000\left(1 + \frac{8}{100}\right)^2 = 10000(1.1664) = 11664 \\ a_3 &= 10000\left(1 + \frac{8}{100}\right)^3 = 10000(1.259712) = 12597.12 \end{aligned}$$
Evaluating consecutive differences:
$$\begin{aligned} a_2 – a_1 &= 11664 – 10800 = 864 \\ a_3 – a_2 &= 12597.12 – 11664 = 933.12 \end{aligned}$$
Since $a_2 – a_1 \neq a_3 – a_2$, the difference between successive terms is not constant due to compounding. Therefore, the list of numbers does not form an AP.
Question 2. Write first four terms of the AP, when the first term $a$ and the common difference $d$ are given as follows:
(i) $a = 10, d = 10$
(ii) $a = -2, d = 0$
(iii) $a = 4, d = -3$
(iv) $a = -1, d = \frac{1}{2}$
(v) $a = -1.25, d = -0.25$ (Page No. 99)
Answer:
The general terms of an AP with first term $a$ and common difference $d$ are given by $a_1 = a$, $a_2 = a + d$, $a_3 = a + 2d$, and $a_4 = a + 3d$.
(i) Given: $a = 10, d = 10$
$$\begin{aligned} a_1 &= a = 10 \\ a_2 &= a + d = 10 + 10 = 20 \\ a_3 &= a + 2d = 10 + 2(10) = 30 \\ a_4 &= a + 3d = 10 + 3(10) = 40 \end{aligned}$$
The first four terms are $10, 20, 30, 40$.
(ii) Given: $a = -2, d = 0$
$$\begin{aligned} a_1 &= a = -2 \\ a_2 &= a + d = -2 + 0 = -2 \\ a_3 &= a + 2d = -2 + 2(0) = -2 \\ a_4 &= a + 3d = -2 + 3(0) = -2 \end{aligned}$$
The first four terms are $-2, -2, -2, -2$.
(iii) Given: $a = 4, d = -3$
$$\begin{aligned} a_1 &= a = 4 \\ a_2 &= a + d = 4 + (-3) = 1 \\ a_3 &= a + 2d = 4 + 2(-3) = 4 – 6 = -2 \\ a_4 &= a + 3d = 4 + 3(-3) = 4 – 9 = -5 \end{aligned}$$
The first four terms are $4, 1, -2, -5$.
(iv) Given: $a = -1, d = \frac{1}{2}$
$$\begin{aligned} a_1 &= a = -1 \\ a_2 &= a + d = -1 + \frac{1}{2} = -\frac{1}{2} \\ a_3 &= a + 2d = -1 + 2\left(\frac{1}{2}\right) = -1 + 1 = 0 \\ a_4 &= a + 3d = -1 + 3\left(\frac{1}{2}\right) = -1 + \frac{3}{2} = \frac{1}{2} \end{aligned}$$
The first four terms are $-1, -\frac{1}{2}, 0, \frac{1}{2}$.
(v) Given: $a = -1.25, d = -0.25$
$$\begin{aligned} a_1 &= a = -1.25 \\ a_2 &= a + d = -1.25 + (-0.25) = -1.50 \\ a_3 &= a + 2d = -1.25 + 2(-0.25) = -1.25 – 0.50 = -1.75 \\ a_4 &= a + 3d = -1.25 + 3(-0.25) = -1.25 – 0.75 = -2.00 \end{aligned}$$
The first four terms are $-1.25, -1.50, -1.75, -2.00$.
Question 3. For the following APs, write the first term and the common difference:
(i) $3, 1, -1, -3, \dots$
(ii) $-5, -1, 3, 7, \dots$
(iii) $\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \dots$
(iv) $0.6, 1.7, 2.8, 3.9, \dots$ (Page No. 100)
Answer:
For an AP sequence $a_1, a_2, a_3, \dots$, the first term is $a = a_1$ and the common difference is $d = a_2 – a_1$.
(i) For the AP $3, 1, -1, -3, \dots$:
$$\begin{aligned} \text{First term } a &= 3 \\ \text{Common difference } d &= a_2 – a_1 = 1 – 3 = -2 \end{aligned}$$
Therefore, $a = 3$ and $d = -2$.
(ii) For the AP $-5, -1, 3, 7, \dots$:
$$\begin{aligned} \text{First term } a &= -5 \\ \text{Common difference } d &= a_2 – a_1 = -1 – (-5) = -1 + 5 = 4 \end{aligned}$$
Therefore, $a = -5$ and $d = 4$.
(iii) For the AP $\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \dots$:
$$\begin{aligned} \text{First term } a &= \frac{1}{3} \\ \text{Common difference } d &= a_2 – a_1 = \frac{5}{3} – \frac{1}{3} = \frac{4}{3} \end{aligned}$$
Therefore, $a = \frac{1}{3}$ and $d = \frac{4}{3}$.
(iv) For the AP $0.6, 1.7, 2.8, 3.9, \dots$:
$$\begin{aligned} \text{First term } a &= 0.6 \\ \text{Common difference } d &= a_2 – a_1 = 1.7 – 0.6 = 1.1 \end{aligned}$$
Therefore, $a = 0.6$ and $d = 1.1$.
Question 4. Which of the following are APs? If they form an AP, find the common difference $d$ and write three more terms.
(i) $2, 4, 8, 16, \dots$
(ii) $2, \frac{5}{2}, 3, \frac{7}{2}, \dots$
(iii) $-1.2, -3.2, -5.2, -7.2, \dots$
(iv) $-10, -6, -2, 2, \dots$
(v) $3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \dots$
(vi) $0.2, 0.22, 0.222, 0.2222, \dots$
(vii) $0, -4, -8, -12, \dots$
(viii) $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \dots$
(ix) $1, 3, 9, 27, \dots$
(x) $a, 2a, 3a, 4a, \dots$
(xi) $a, a^2, a^3, a^4, \dots$
(xii) $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots$
(xiii) $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots$
(xiv) $1^2, 3^2, 5^2, 7^2, \dots$
(xv) $1^2, 5^2, 7^2, 73, \dots$ (Page No. 100)
Answer:
A list of numbers forms an AP if $a_{k+1} – a_k$ remains constant for all consecutive pairs. If it forms an AP, the next three terms are $a_5 = a_4 + d$, $a_6 = a_5 + d$, and $a_7 = a_6 + d$.
(i) $2, 4, 8, 16, \dots$
$$\begin{aligned} a_2 – a_1 &= 4 – 2 = 2 \\ a_3 – a_2 &= 8 – 4 = 4 \end{aligned}$$
Since $a_2 – a_1 \neq a_3 – a_2$, it does not form an AP.
(ii) $2, \frac{5}{2}, 3, \frac{7}{2}, \dots$
$$\begin{aligned} a_2 – a_1 &= \frac{5}{2} – 2 = \frac{1}{2} \\ a_3 – a_2 &= 3 – \frac{5}{2} = \frac{1}{2} \\ a_4 – a_3 &= \frac{7}{2} – 3 = \frac{1}{2} \end{aligned}$$
Since the common difference $d = \frac{1}{2}$ is constant, it forms an AP.
The next three terms are:
$$\begin{aligned} a_5 &= \frac{7}{2} + \frac{1}{2} = \frac{8}{2} = 4 \\ a_6 &= 4 + \frac{1}{2} = \frac{9}{2} \\ a_7 &= \frac{9}{2} + \frac{1}{2} = \frac{10}{2} = 5 \end{aligned}$$
Common difference $d = \frac{1}{2}$; Next three terms: $4, \frac{9}{2}, 5$.
(iii) $-1.2, -3.2, -5.2, -7.2, \dots$
$$\begin{aligned} a_2 – a_1 &= -3.2 – (-1.2) = -2.0 \\ a_3 – a_2 &= -5.2 – (-3.2) = -2.0 \\ a_4 – a_3 &= -7.2 – (-5.2) = -2.0 \end{aligned}$$
Since $d = -2.0$ is constant, it forms an AP.
The next three terms are:
$$\begin{aligned} a_5 &= -7.2 + (-2.0) = -9.2 \\ a_6 &= -9.2 + (-2.0) = -11.2 \\ a_7 &= -11.2 + (-2.0) = -13.2 \end{aligned}$$
Common difference $d = -2$; Next three terms: $-9.2, -11.2, -13.2$.
(iv) $-10, -6, -2, 2, \dots$
$$\begin{aligned} a_2 – a_1 &= -6 – (-10) = 4 \\ a_3 – a_2 &= -2 – (-6) = 4 \\ a_4 – a_3 &= 2 – (-2) = 4 \end{aligned}$$
Since $d = 4$ is constant, it forms an AP.
The next three terms are:
$$\begin{aligned} a_5 &= 2 + 4 = 6 \\ a_6 &= 6 + 4 = 10 \\ a_7 &= 10 + 4 = 14 \end{aligned}$$
Common difference $d = 4$; Next three terms: $6, 10, 14$.
(v) $3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \dots$
$$\begin{aligned} a_2 – a_1 &= (3 + \sqrt{2}) – 3 = \sqrt{2} \\ a_3 – a_2 &= (3 + 2\sqrt{2}) – (3 + \sqrt{2}) = \sqrt{2} \\ a_4 – a_3 &= (3 + 3\sqrt{2}) – (3 + 2\sqrt{2}) = \sqrt{2} \end{aligned}$$
Since $d = \sqrt{2}$ is constant, it forms an AP.
The next three terms are:
$$\begin{aligned} a_5 &= (3 + 3\sqrt{2}) + \sqrt{2} = 3 + 4\sqrt{2} \\ a_6 &= (3 + 4\sqrt{2}) + \sqrt{2} = 3 + 5\sqrt{2} \\ a_7 &= (3 + 5\sqrt{2}) + \sqrt{2} = 3 + 6\sqrt{2} \end{aligned}$$
Common difference $d = \sqrt{2}$; Next three terms: $3+4\sqrt{2}, 3+5\sqrt{2}, 3+6\sqrt{2}$.
(vi) $0.2, 0.22, 0.222, 0.2222, \dots$
$$\begin{aligned} a_2 – a_1 &= 0.22 – 0.2 = 0.02 \\ a_3 – a_2 &= 0.222 – 0.22 = 0.002 \end{aligned}$$
Since $a_2 – a_1 \neq a_3 – a_2$, it does not form an AP.
(vii) $0, -4, -8, -12, \dots$
$$\begin{aligned} a_2 – a_1 &= -4 – 0 = -4 \\ a_3 – a_2 &= -8 – (-4) = -4 \\ a_4 – a_3 &= -12 – (-8) = -4 \end{aligned}$$
Since $d = -4$ is constant, it forms an AP.
The next three terms are:
$$\begin{aligned} a_5 &= -12 + (-4) = -16 \\ a_6 &= -16 + (-4) = -20 \\ a_7 &= -20 + (-4) = -24 \end{aligned}$$
Common difference $d = -4$; Next three terms: $-16, -20, -24$.
(viii) $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \dots$
$$\begin{aligned} a_2 – a_1 &= -\frac{1}{2} – \left(-\frac{1}{2}\right) = 0 \\ a_3 – a_2 &= -\frac{1}{2} – \left(-\frac{1}{2}\right) = 0 \\ a_4 – a_3 &= -\frac{1}{2} – \left(-\frac{1}{2}\right) = 0 \end{aligned}$$
Since $d = 0$ is constant, it forms an AP.
The next three terms are:
$$\begin{aligned} a_5 &= -\frac{1}{2} + 0 = -\frac{1}{2} \\ a_6 &= -\frac{1}{2} + 0 = -\frac{1}{2} \\ a_7 &= -\frac{1}{2} + 0 = -\frac{1}{2} \end{aligned}$$
Common difference $d = 0$; Next three terms: $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}$.
(ix) $1, 3, 9, 27, \dots$
$$\begin{aligned} a_2 – a_1 &= 3 – 1 = 2 \\ a_3 – a_2 &= 9 – 3 = 6 \end{aligned}$$
Since $a_2 – a_1 \neq a_3 – a_2$, it does not form an AP.
(x) $a, 2a, 3a, 4a, \dots$
$$\begin{aligned} a_2 – a_1 &= 2a – a = a \\ a_3 – a_2 &= 3a – 2a = a \\ a_4 – a_3 &= 4a – 3a = a \end{aligned}$$
Since $d = a$ is constant, it forms an AP.
The next three terms are:
$$\begin{aligned}
a_5 &= 4a + a = 5a \
a_6 &= 5a + a = 6a \## NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions
1. SEO Strategy Introduction & Chapter Master Overview
Chapter 5, Arithmetic Progressions (AP), is an essential component of the CBSE Class 10 Mathematics syllabus under Unit II: Algebra. In the official CBSE Board blueprint, this chapter carries an average weightage of 4 to 6 marks, distributed across objective Multiple Choice Questions (1 mark), Short Answer Questions (2 or 3 marks), and high-weightage Case Study / Long Answer Questions (4 to 5 marks). Mastering Arithmetic Progressions builds the mathematical foundation for analyzing discrete linear models, financial sequences, series summations, and higher-level progression concepts in Class 11 and 12.
An Arithmetic Progression is formally defined as a sequence of numbers in which the difference between any two consecutive terms remains constant throughout. This fixed numerical difference is termed the common difference ($d$), and the starting value is denoted as the first term ($a$). CBSE marking schemes prioritize clear algebraic definitions, identification of given parameters, explicit statement of standard formulas, step-by-step substitution, and boxed final answers with appropriate physical units.
Plaintext
+----------------------------------------------------------------------------------------------------+
| MASTER SUMMARY: ARITHMETIC PROGRESSIONS |
+-----------------------------------+-----------------------------------+----------------------------+
| Concept / Parameter | Mathematical Formula / Condition | Key Scoring Notes |
+-----------------------------------+-----------------------------------+----------------------------+
| General Form of an AP | a, a + d, a + 2d, a + 3d, ... | 'a' = first term |
| | | 'd' = common difference |
+-----------------------------------+-----------------------------------+----------------------------+
| Common Difference (d) | d = a_{k+1} - a_k | Constant; can be +, -, 0 |
+-----------------------------------+-----------------------------------+----------------------------+
| n-th Term (General Term) | a_n = a + (n - 1)d | n in Natural Numbers (N) |
+-----------------------------------+-----------------------------------+----------------------------+
| n-th Term from End (Last = l) | a_n' = l - (n - 1)d | Reversing AP gives d' = -d |
+-----------------------------------+-----------------------------------+----------------------------+
| Sum of First n Terms (Standard) | S_n = (n / 2) * [2a + (n - 1)d] | Used when 'd' is known |
+-----------------------------------+-----------------------------------+----------------------------+
| Sum of First n Terms (Last Term) | S_n = (n / 2) * (a + l) | Used when 'l' is known |
+-----------------------------------+-----------------------------------+----------------------------+
| Term via Sums Relation | a_n = S_n - S_{n-1} | Valid for all n >= 2 |
+-----------------------------------+-----------------------------------+----------------------------+
| 3 Symmetric Terms in AP | (a - d), a, (a + d) | Common difference = d |
+-----------------------------------+-----------------------------------+----------------------------+
| 4 Symmetric Terms in AP | (a - 3d), (a - d), (a + d), | Common difference = 2d |
| | (a + 3d) | |
+-----------------------------------+-----------------------------------+----------------------------+
| Arithmetic Mean (AM) of a and c | b = (a + c) / 2 | Condition: 2b = a + c |
+-----------------------------------+-----------------------------------+----------------------------+
2. Exercise 5.1 Solutions (Page No. 99-100)
Question 1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes $\frac{1}{4}$ of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.
(iv) The amount of money in the account every year, when ₹ 10,000 is deposited at compound interest at $8\%$ per annum. (Page No. 99)
Answer:
(i) Let $a_n$ represent the total taxi fare for $n\text{ km}$.
$$\begin{aligned} a_1 &= 15 \\ a_2 &= 15 + 8 = 23 \\ a_3 &= 23 + 8 = 31 \\ a_4 &= 31 + 8 = 39 \end{aligned}$$
Calculating consecutive differences:
$$\begin{aligned} a_2 – a_1 &= 23 – 15 = 8 \\ a_3 – a_2 &= 31 – 23 = 8 \\ a_4 – a_3 &= 39 – 31 = 8 \end{aligned}$$
Since the difference $a_{k+1} – a_k = 8$ is constant for all $k$, the list of numbers $15, 23, 31, 39, \dots$ forms an Arithmetic Progression with $a = 15$ and $d = 8$.
(ii) Let the initial volume of gas present in the cylinder be $V$.
$$\begin{aligned} a_1 &= V \\ a_2 &= V – \frac{1}{4}V = \frac{3}{4}V \\ a_3 &= \frac{3}{4}V – \frac{1}{4}\left(\frac{3}{4}V\right) = \left(\frac{3}{4}\right)^2 V = \frac{9}{16}V \\ a_4 &= \frac{9}{16}V – \frac{1}{4}\left(\frac{9}{16}V\right) = \left(\frac{3}{4}\right)^3 V = \frac{27}{64}V \end{aligned}$$
Testing consecutive differences:
$$\begin{aligned} a_2 – a_1 &= \frac{3}{4}V – V = -\frac{1}{4}V \\ a_3 – a_2 &= \frac{9}{16}V – \frac{3}{4}V = -\frac{3}{16}V \end{aligned}$$
Because $a_2 – a_1 \neq a_3 – a_2$, the difference between successive terms is not constant. Hence, this situation does not form an AP.
(iii) Let $a_n$ denote the cumulative cost of digging $n$ metres.
$$\begin{aligned} a_1 &= 150 \\ a_2 &= 150 + 50 = 200 \\ a_3 &= 200 + 50 = 250 \\ a_4 &= 250 + 50 = 300 \end{aligned}$$
Evaluating differences:
$$\begin{aligned} a_2 – a_1 &= 200 – 150 = 50 \\ a_3 – a_2 &= 250 – 200 = 50 \\ a_4 – a_3 &= 300 – 250 = 50 \end{aligned}$$
Since $a_{k+1} – a_k = 50$ is constant, the given sequence $150, 200, 250, 300, \dots$ forms an AP with $a = 150$ and $d = 50$.
(iv) Let $P = \text{₹ } 10000$ and $r = 8\%$ compounded annually. The amount $A_n$ at year $n$ is $P\left(1 + \frac{r}{100}\right)^n$.
$$\begin{aligned} a_1 &= 10000\left(1 + \frac{8}{100}\right)^1 = 10800 \\ a_2 &= 10000\left(1.08\right)^2 = 11664 \\ a_3 &= 10000\left(1.08\right)^3 = 12597.12 \end{aligned}$$
Differences:
$$\begin{aligned} a_2 – a_1 &= 11664 – 10800 = 864 \\ a_3 – a_2 &= 12597.12 – 11664 = 933.12 \end{aligned}$$
Since $a_2 – a_1 \neq a_3 – a_2$, this sequence does not form an AP.
Question 2. Write first four terms of the AP, when the first term $a$ and the common difference $d$ are given as follows:
(i) $a = 10, d = 10$
(ii) $a = -2, d = 0$
(iii) $a = 4, d = -3$
(iv) $a = -1, d = \frac{1}{2}$
(v) $a = -1.25, d = -0.25$ (Page No. 99)
Answer:
Applying $a_1 = a$, $a_2 = a + d$, $a_3 = a + 2d$, and $a_4 = a + 3d$:
- (i) $a = 10, d = 10$:$a_1 = 10$, $a_2 = 10+10 = 20$, $a_3 = 20+10 = 30$, $a_4 = 30+10 = 40$.First four terms: $10, 20, 30, 40$.
- (ii) $a = -2, d = 0$:$a_1 = -2$, $a_2 = -2+0 = -2$, $a_3 = -2+0 = -2$, $a_4 = -2+0 = -2$.First four terms: $-2, -2, -2, -2$.
- (iii) $a = 4, d = -3$:$a_1 = 4$, $a_2 = 4 + (-3) = 1$, $a_3 = 1 + (-3) = -2$, $a_4 = -2 + (-3) = -5$.First four terms: $4, 1, -2, -5$.
- (iv) $a = -1, d = \frac{1}{2}$:$a_1 = -1$, $a_2 = -1 + \frac{1}{2} = -\frac{1}{2}$, $a_3 = -\frac{1}{2} + \frac{1}{2} = 0$, $a_4 = 0 + \frac{1}{2} = \frac{1}{2}$.First four terms: $-1, -\frac{1}{2}, 0, \frac{1}{2}$.
- (v) $a = -1.25, d = -0.25$:$a_1 = -1.25$, $a_2 = -1.25 + (-0.25) = -1.50$, $a_3 = -1.50 + (-0.25) = -1.75$, $a_4 = -1.75 + (-0.25) = -2.00$.First four terms: $-1.25, -1.50, -1.75, -2.00$.
Question 3. For the following APs, write the first term and the common difference:
(i) $3, 1, -1, -3, \dots$
(ii) $-5, -1, 3, 7, \dots$
(iii) $\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \dots$
(iv) $0.6, 1.7, 2.8, 3.9, \dots$ (Page No. 100)
Answer:
- (i) $a = a_1 = 3$; $d = 1 – 3 = -2$. Thus, $a = 3, d = -2$.
- (ii) $a = a_1 = -5$; $d = -1 – (-5) = 4$. Thus, $a = -5, d = 4$.
- (iii) $a = a_1 = \frac{1}{3}$; $d = \frac{5}{3} – \frac{1}{3} = \frac{4}{3}$. Thus, $a = \frac{1}{3}, d = \frac{4}{3}$.
- (iv) $a = a_1 = 0.6$; $d = 1.7 – 0.6 = 1.1$. Thus, $a = 0.6, d = 1.1$.
Question 4. Which of the following are APs? If they form an AP, find the common difference $d$ and write three more terms.
(i) $2, 4, 8, 16, \dots$
(ii) $2, \frac{5}{2}, 3, \frac{7}{2}, \dots$
(iii) $-1.2, -3.2, -5.2, -7.2, \dots$
(iv) $-10, -6, -2, 2, \dots$
(v) $3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \dots$
(vi) $0.2, 0.22, 0.222, 0.2222, \dots$
(vii) $0, -4, -8, -12, \dots$
(viii) $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \dots$
(ix) $1, 3, 9, 27, \dots$
(x) $a, 2a, 3a, 4a, \dots$
(xi) $a, a^2, a^3, a^4, \dots$
(xii) $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots$
(xiii) $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots$
(xiv) $1^2, 3^2, 5^2, 7^2, \dots$
(xv) $1^2, 5^2, 7^2, 73, \dots$ (Page No. 100)
Answer:
- (i) $4 – 2 = 2 \neq 8 – 4 = 4$. Not an AP.
- (ii) $\frac{5}{2} – 2 = \frac{1}{2}$, $3 – \frac{5}{2} = \frac{1}{2}$, $\frac{7}{2} – 3 = \frac{1}{2}$. Constant difference $d = \frac{1}{2}$. Forms an AP.Next terms: $a_5 = \frac{7}{2}+\frac{1}{2}=4$, $a_6 = 4+\frac{1}{2}=\frac{9}{2}$, $a_7 = \frac{9}{2}+\frac{1}{2}=5$. $d = \frac{1}{2}$; Terms: $4, \frac{9}{2}, 5$.
- (iii) Differences: $-3.2 – (-1.2) = -2$, $-5.2 – (-3.2) = -2$. Constant $d = -2$. Forms an AP.Next terms: $-9.2, -11.2, -13.2$. $d = -2$; Terms: $-9.2, -11.2, -13.2$.
- (iv) Differences: $-6 – (-10) = 4$, $-2 – (-6) = 4$, $2 – (-2) = 4$. Constant $d = 4$. Forms an AP.Next terms: $6, 10, 14$. $d = 4$; Terms: $6, 10, 14$.
- (v) Differences: $(3+\sqrt{2})-3 = \sqrt{2}$, $(3+2\sqrt{2})-(3+\sqrt{2}) = \sqrt{2}$. Constant $d = \sqrt{2}$. Forms an AP.Next terms: $3+4\sqrt{2}, 3+5\sqrt{2}, 3+6\sqrt{2}$. $d = \sqrt{2}$; Terms: $3+4\sqrt{2}, 3+5\sqrt{2}, 3+6\sqrt{2}$.
- (vi) $0.22 – 0.2 = 0.02 \neq 0.222 – 0.22 = 0.002$. Not an AP.
- (vii) Differences: $-4 – 0 = -4$, $-8 – (-4) = -4$. Constant $d = -4$. Forms an AP.Next terms: $-16, -20, -24$. $d = -4$; Terms: $-16, -20, -24$.
- (viii) Differences: $-\frac{1}{2} – (-\frac{1}{2}) = 0$. Constant $d = 0$. Forms an AP.Next terms: $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}$. $d = 0$; Terms: $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}$.
- (ix) $3 – 1 = 2 \neq 9 – 3 = 6$. Not an AP.
- (x) Differences: $2a – a = a$, $3a – 2a = a$, $4a – 3a = a$. Constant $d = a$. Forms an AP.Next terms: $5a, 6a, 7a$. $d = a$; Terms: $5a, 6a, 7a$.
- (xi) $a^2 – a = a(a – 1) \neq a^3 – a^2 = a^2(a – 1)$ (for $a \neq 0, 1$). Not an AP.
- (xii) Rewriting: $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \dots$Differences: $2\sqrt{2}-\sqrt{2}=\sqrt{2}$, $3\sqrt{2}-2\sqrt{2}=\sqrt{2}$. Constant $d = \sqrt{2}$. Forms an AP.Next terms: $5\sqrt{2} = \sqrt{50}$, $6\sqrt{2} = \sqrt{72}$, $7\sqrt{2} = \sqrt{98}$. $d = \sqrt{2}$; Terms: $\sqrt{50}, \sqrt{72}, \sqrt{98}$.
- (xiii) $\sqrt{6} – \sqrt{3} \neq \sqrt{9} – \sqrt{6} = 3 – \sqrt{6}$. Not an AP.
- (xiv) $1, 9, 25, 49, \dots \implies 9 – 1 = 8 \neq 25 – 9 = 16$. Not an AP.
- (xv) $1, 25, 49, 73, \dots \implies 25 – 1 = 24$, $49 – 25 = 24$, $73 – 49 = 24$. Constant $d = 24$. Forms an AP.Next terms: $73 + 24 = 97$, $97 + 24 = 121$, $121 + 24 = 145$. $d = 24$; Terms: $97, 121, 145$.
3. Exercise 5.2 Solutions (Page No. 105-107)
Question 1. Fill in the blanks in the following table, given that $a$ is the first term, $d$ the common difference and $a_n$ the $n^{\text{th}}$ term of the AP:
| S.No. | a | d | n | an |
| (i) | $7$ | $3$ | $8$ | $\dots$ |
| (ii) | $-18$ | $\dots$ | $10$ | $0$ |
| (iii) | $\dots$ | $-3$ | $18$ | $-5$ |
| (iv) | $-18.9$ | $2.5$ | $\dots$ | $3.6$ |
| (v) | $3.5$ | $0$ | $105$ | $\dots$ |
Answer:
The general formula is $a_n = a + (n – 1)d$.
- (i) Given $a = 7, d = 3, n = 8$:$$a_8 = 7 + (8 – 1)3 = 7 + 21 = 28$$Blank value: $a_n = 28$.
- (ii) Given $a = -18, n = 10, a_n = 0$:$$0 = -18 + (10 – 1)d \implies 9d = 18 \implies d = 2$$Blank value: $d = 2$.
- (iii) Given $d = -3, n = 18, a_n = -5$:$$-5 = a + (18 – 1)(-3) \implies -5 = a – 51 \implies a = 46$$Blank value: $a = 46$.
- (iv) Given $a = -18.9, d = 2.5, a_n = 3.6$:$$3.6 = -18.9 + (n – 1)2.5 \implies 2.5(n – 1) = 22.5 \implies n – 1 = 9 \implies n = 10$$Blank value: $n = 10$.
- (v) Given $a = 3.5, d = 0, n = 105$:$$a_{105} = 3.5 + (105 – 1)0 = 3.5$$Blank value: $a_n = 3.5$.
Question 2. Choose the correct choice in the following and justify:
(i) $30^{\text{th}}$ term of the AP: $10, 7, 4, \dots$, is
(A) $97$
(B) $77$
(C) $-77$
(D) $-87$
(ii) $11^{\text{th}}$ term of the AP: $-3, -\frac{1}{2}, 2, \dots$, is
(A) $28$
(B) $22$
(C) $-38$
(D) $-46\frac{1}{2}$ (Page No. 106)
Answer:
(i) Here, $a = 10$, $d = 7 – 10 = -3$, and $n = 30$.
$$a_{30} = 10 + (30 – 1)(-3) = 10 + 29(-3) = 10 – 87 = -77$$
Correct Option: (C) $-77$.
(ii) Here, $a = -3$, $d = -\frac{1}{2} – (-3) = \frac{5}{2}$, and $n = 11$.
$$a_{11} = -3 + (11 – 1)\left(\frac{5}{2}\right) = -3 + 10\left(\frac{5}{2}\right) = -3 + 25 = 22$$
Correct Option: (B) $22$.
Question 3. In the following APs, find the missing terms in the boxes:
(i) $2, \; \square, \; 26$
(ii) $\square, \; 13, \; \square, \; 3$
(iii) $5, \; \square, \; \square, \; 9\frac{1}{2}$
(iv) $-4, \; \square, \; \square, \; \square, \; \square, \; 6$
(v) $\square, \; 38, \; \square, \; \square, \; \square, \; -22$ (Page No. 106)
Answer:
- (i) $2, \; \square, \; 26$:Let missing term be $x$. Since $2, x, 26$ are in AP, $x = \frac{2 + 26}{2} = 14$.Missing term: $14$.
- (ii) $\square, \; 13, \; \square, \; 3$:$a_2 = a + d = 13$ and $a_4 = a + 3d = 3$.Subtracting gives $2d = -10 \implies d = -5$.Then $a = 13 – (-5) = 18$, and $a_3 = 13 + (-5) = 8$.Missing terms: $18$ and $8$.
- (iii) $5, \; \square, \; \square, \; \frac{19}{2}$:$a = 5$, $a_4 = a + 3d = \frac{19}{2} \implies 3d = \frac{19}{2} – 5 = \frac{9}{2} \implies d = \frac{3}{2}$.$a_2 = 5 + \frac{3}{2} = \frac{13}{2} = 6\frac{1}{2}$; $a_3 = \frac{13}{2} + \frac{3}{2} = 8$.Missing terms: $6\frac{1}{2}$ and $8$.
- (iv) $-4, \; \square, \; \square, \; \square, \; \square, \; 6$:$a = -4$, $a_6 = a + 5d = 6 \implies -4 + 5d = 6 \implies 5d = 10 \implies d = 2$.$a_2 = -2$, $a_3 = 0$, $a_4 = 2$, $a_5 = 4$.Missing terms: $-2, 0, 2, 4$.
- (v) $\square, \; 38, \; \square, \; \square, \; \square, \; -22$:$a_2 = a + d = 38$ and $a_6 = a + 5d = -22$.Subtracting gives $4d = -60 \implies d = -15$.$a = 38 – (-15) = 53$; $a_3 = 38 – 15 = 23$; $a_4 = 23 – 15 = 8$; $a_5 = 8 – 15 = -7$.Missing terms: $53, 23, 8, -7$.
Question 4. Which term of the AP: $3, 8, 13, 18, \dots$, is $78$? (Page No. 106)
Answer:
Given AP: $a = 3$, $d = 8 – 3 = 5$. Let $a_n = 78$.
$$\begin{aligned} a_n &= a + (n – 1)d \\ 78 &= 3 + (n – 1)5 \\ 75 &= 5(n – 1) \\ n – 1 &= 15 \implies n = 16 \end{aligned}$$
Thus, the $16^{\text{th}}$ term of the AP is $78$.
Question 5. Find the number of terms in each of the following APs:
(i) $7, 13, 19, \dots, 205$
(ii) $18, 15\frac{1}{2}, 13, \dots, -47$ (Page No. 106)
Answer:
(i) $a = 7, d = 6, a_n = 205$:
$$205 = 7 + (n – 1)6 \implies 198 = 6(n – 1) \implies n – 1 = 33 \implies n = 34$$
There are $34$ terms.
(ii) $a = 18, d = \frac{31}{2} – 18 = -\frac{5}{2}, a_n = -47$:
$$-47 = 18 + (n – 1)\left(-\frac{5}{2}\right) \implies -65 = (n – 1)\left(-\frac{5}{2}\right) \implies n – 1 = 26 \implies n = 27$$
There are $27$ terms.
Question 6. Check whether $-150$ is a term of the AP: $11, 8, 5, 2, \dots$ (Page No. 106)
Answer:
For the given AP, $a = 11$ and $d = 8 – 11 = -3$. Assume $a_n = -150$.
$$\begin{aligned} -150 &= 11 + (n – 1)(-3) \\ -161 &= -3(n – 1) \\ n – 1 &= \frac{161}{3} \implies n = \frac{164}{3} = 54\frac{2}{3} \end{aligned}$$
Since $n$ must be a positive integer ($n \in \mathbb{N}$), $-150$ is not a term of this AP.
Question 7. Find the $31^{\text{st}}$ term of an AP whose $11^{\text{th}}$ term is $38$ and the $16^{\text{th}}$ term is $73$. (Page No. 106)
Answer:
Given $a_{11} = a + 10d = 38$ and $a_{16} = a + 15d = 73$.
$$\begin{aligned} (a + 15d) – (a + 10d) &= 73 – 38 \\ 5d &= 35 \implies d = 7 \end{aligned}$$
Substituting $d = 7$ into $a + 10(7) = 38 \implies a = 38 – 70 = -32$.
$$a_{31} = a + 30d = -32 + 30(7) = -32 + 210 = 178$$
The $31^{\text{st}}$ term is $178$.
Question 8. An AP consists of $50$ terms of which $3^{\text{rd}}$ term is $12$ and the last term is $106$. Find the $29^{\text{th}}$ term. (Page No. 106)
Answer:
Given $n = 50$, $a_3 = a + 2d = 12$, and $a_{50} = a + 49d = 106$.
$$47d = 106 – 12 = 94 \implies d = 2$$
Then $a = 12 – 2(2) = 8$.
$$a_{29} = a + 28d = 8 + 28(2) = 8 + 56 = 64$$
The $29^{\text{th}}$ term is $64$.
Question 9. If the $3^{\text{rd}}$ and the $9^{\text{th}}$ terms of an AP are $4$ and $-8$ respectively, which term of this AP is zero? (Page No. 106)
Answer:
Given $a_3 = a + 2d = 4$ and $a_9 = a + 8d = -8$.
$$6d = -8 – 4 = -12 \implies d = -2$$
Thus $a + 2(-2) = 4 \implies a = 8$.
Setting $a_n = 0$:
$$0 = 8 + (n – 1)(-2) \implies 2(n – 1) = 8 \implies n = 5$$
The $5^{\text{th}}$ term is zero.
Question 10. The $17^{\text{th}}$ term of an AP exceeds its $10^{\text{th}}$ term by $7$. Find the common difference. (Page No. 106)
Answer:
According to the question:
$$\begin{aligned} a_{17} – a_{10} &= 7 \\ [a + 16d] – [a + 9d] &= 7 \\ 7d &= 7 \implies d = 1 \end{aligned}$$
The common difference is $d = 1$.
Question 11. Which term of the AP: $3, 15, 27, 39, \dots$ will be $132$ more than its $54^{\text{th}}$ term? (Page No. 106)
Answer:
Here, $a = 3$ and $d = 15 – 3 = 12$.
$$a_{54} = 3 + 53(12) = 3 + 636 = 639$$
Let the required term be $a_n = 639 + 132 = 771$.
$$771 = 3 + (n – 1)12 \implies 12(n – 1) = 768 \implies n – 1 = 64 \implies n = 65$$
The $65^{\text{th}}$ term is $132$ more than its $54^{\text{th}}$ term.
Question 12. Two APs have the same common difference. The difference between their $100^{\text{th}}$ terms is $100$, what is the difference between their $1000^{\text{th}}$ terms? (Page No. 106)
Answer:
Let the first terms of the two APs be $a$ and $A$, with identical common difference $d$.
The $n^{\text{th}}$ terms are $a_n = a + (n – 1)d$ and $A_n = A + (n – 1)d$.
Difference between $n^{\text{th}}$ terms:
$$a_n – A_n = [a + (n – 1)d] – [A + (n – 1)d] = a – A$$
Given $a_{100} – A_{100} = 100 \implies a – A = 100$.
Therefore, for $n = 1000$:
$$a_{1000} – A_{1000} = a – A = 100$$
The difference between their $1000^{\text{th}}$ terms is $100$.
Question 13. How many three-digit numbers are divisible by $7$? (Page No. 106)
Answer:
The smallest 3-digit number divisible by $7$ is $105$, and the largest is $994$.
The sequence is $105, 112, 119, \dots, 994$, which forms an AP with $a = 105, d = 7, a_n = 994$.
$$994 = 105 + (n – 1)7 \implies 7(n – 1) = 889 \implies n – 1 = 127 \implies n = 128$$
There are $128$ three-digit numbers divisible by $7$.
Question 14. How many multiples of $4$ lie between $10$ and $250$? (Page No. 106)
Answer:
The multiples of $4$ in this range are $12, 16, 20, \dots, 248$.
Here $a = 12, d = 4, a_n = 248$.
$$248 = 12 + (n – 1)4 \implies 4(n – 1) = 236 \implies n – 1 = 59 \implies n = 60$$
There are $60$ multiples of $4$ between $10$ and $250$.
Question 15. For what value of $n$, are the $n^{\text{th}}$ terms of two APs: $63, 65, 67, \dots$ and $3, 10, 17, \dots$ equal? (Page No. 106)
Answer:
For the $1^{\text{st}}$ AP ($63, 65, 67, \dots$): $a_1 = 63, d_1 = 2 \implies a_n = 63 + (n – 1)2 = 2n + 61$.
For the $2^{\text{nd}}$ AP ($3, 10, 17, \dots$): $a_2 = 3, d_2 = 7 \implies A_n = 3 + (n – 1)7 = 7n – 4$.
Equating the two general terms:
$$2n + 61 = 7n – 4 \implies 5n = 65 \implies n = 13$$
The $n^{\text{th}}$ terms are equal for $n = 13$.
Question 16. Determine the AP whose third term is $16$ and the $7^{\text{th}}$ term exceeds the $5^{\text{th}}$ term by $12$. (Page No. 106)
Answer:
Given $a_3 = a + 2d = 16$ and $a_7 – a_5 = 12$.
$$(a + 6d) – (a + 4d) = 12 \implies 2d = 12 \implies d = 6$$
Substituting $d = 6$: $a + 2(6) = 16 \implies a = 4$.
The AP is $a, a + d, a + 2d, \dots \implies$ $4, 10, 16, 22, \dots$
Question 17. Find the $20^{\text{th}}$ term from the last term of the AP: $3, 8, 13, \dots, 253$. (Page No. 107)
Answer:
Method: Reversing the AP makes the last term the first term, with the sign of the common difference inverted.
Reversed AP: $253, 248, 243, \dots, 3$, where $a = 253$ and $d = -5$.
$$a_{20} = 253 + (20 – 1)(-5) = 253 + 19(-5) = 253 – 95 = 158$$
The $20^{\text{th}}$ term from the end is $158$.
Question 18. The sum of the $4^{\text{th}}$ and $8^{\text{th}}$ terms of an AP is $24$ and the sum of the $6^{\text{th}}$ and $10^{\text{th}}$ terms is $44$. Find the first three terms of the AP. (Page No. 107)
Answer:
From the first condition:
$$a_4 + a_8 = 24 \implies (a + 3d) + (a + 7d) = 24 \implies 2a + 10d = 24 \implies a + 5d = 12 \quad \text{— (1)}$$
From the second condition:
$$a_6 + a_{10} = 44 \implies (a + 5d) + (a + 9d) = 44 \implies 2a + 14d = 44 \implies a + 7d = 22 \quad \text{— (2)}$$
Subtracting (1) from (2):
$$2d = 10 \implies d = 5$$
Substituting $d = 5$ into (1): $a + 5(5) = 12 \implies a = -13$.
First three terms: $a_1 = -13$, $a_2 = -13 + 5 = -8$, $a_3 = -8 + 5 = -3$.
The first three terms are $-13, -8, -3$.
Question 19. Subba Rao started work in $1995$ at an annual salary of ₹ $5000$ and received an increment of ₹ $200$ each year. In which year did his income reach ₹ $7000$? (Page No. 107)
Answer:
The annual salaries form an AP: $5000, 5200, 5400, \dots, 7000$, with $a = 5000, d = 200, a_n = 7000$.
$$7000 = 5000 + (n – 1)200 \implies 200(n – 1) = 2000 \implies n – 1 = 10 \implies n = 11$$
The income reached ₹ $7000$ in the $11^{\text{th}}$ year, which corresponds to $1995 + (11 – 1) =$ $2005$.
Question 20. Ramkali saved ₹ $5$ in the first week of a year and then increased her weekly savings by ₹ $1.75$. If in the $n^{\text{th}}$ week, her weekly savings become ₹ $20.75$, find $n$. (Page No. 107)
Answer:
Here, $a = 5$, $d = 1.75$, and $a_n = 20.75$.
$$\begin{aligned} 20.75 &= 5 + (n – 1)(1.75) \\ 15.75 &= 1.75(n – 1) \\ n – 1 &= \frac{15.75}{1.75} = 9 \implies n = 10 \end{aligned}$$
Thus, $n = 10$.
4. Exercise 5.3 Solutions (Page No. 112-114)
Question 1. Find the sum of the following APs:
(i) $2, 7, 12, \dots$, to $10$ terms.
(ii) $-37, -33, -29, \dots$, to $12$ terms.
(iii) $0.6, 1.7, 2.8, \dots$, to $100$ terms.
(iv) $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \dots$, to $11$ terms. (Page No. 112)
Answer:
The sum formula is $S_n = \frac{n}{2}[2a + (n – 1)d]$.
- (i) $a = 2, d = 5, n = 10$:$$S_{10} = \frac{10}{2}[2(2) + (10 – 1)5] = 5[4 + 45] = 5(49) = 245$$Sum: $245$.
- (ii) $a = -37, d = 4, n = 12$:$$S_{12} = \frac{12}{2}[2(-37) + (12 – 1)4] = 6[-74 + 44] = 6(-30) = -180$$Sum: $-180$.
- (iii) $a = 0.6, d = 1.1, n = 100$:$$S_{100} = \frac{100}{2}[2(0.6) + 99(1.1)] = 50[1.2 + 108.9] = 50(110.1) = 5505$$Sum: $5505$.
- (iv) $a = \frac{1}{15}$, $d = \frac{1}{12} – \frac{1}{15} = \frac{5 – 4}{60} = \frac{1}{60}$, $n = 11$:$$S_{11} = \frac{11}{2}\left[2\left(\frac{1}{15}\right) + 10\left(\frac{1}{60}\right)\right] = \frac{11}{2}\left[\frac{2}{15} + \frac{1}{6}\right] = \frac{11}{2}\left[\frac{4 + 5}{30}\right] = \frac{11}{2}\left(\frac{9}{30}\right) = \frac{33}{20}$$Sum: $\frac{33}{20}$.
Question 2. Find the sums given below:
(i) $7 + 10\frac{1}{2} + 14 + \dots + 84$
(ii) $34 + 32 + 30 + \dots + 10$
(iii) $-5 + (-8) + (-11) + \dots + (-230)$ (Page No. 112)
Answer:
(i) $a = 7, d = \frac{7}{2}, l = 84$.
Finding $n$:
$$84 = 7 + (n – 1)\left(\frac{7}{2}\right) \implies 77 = (n – 1)\left(\frac{7}{2}\right) \implies n – 1 = 22 \implies n = 23$$
Sum:
$$S_{23} = \frac{23}{2}(7 + 84) = \frac{23 \times 91}{2} = \frac{2093}{2} = 1046\frac{1}{2}$$
Sum: $1046\frac{1}{2}$.
(ii) $a = 34, d = -2, l = 10$.
Finding $n$:
$$10 = 34 + (n – 1)(-2) \implies -24 = -2(n – 1) \implies n – 1 = 12 \implies n = 13$$
Sum:
$$S_{13} = \frac{13}{2}(34 + 10) = \frac{13 \times 44}{2} = 286$$
Sum: $286$.
(iii) $a = -5, d = -3, l = -230$.
Finding $n$:
$$-230 = -5 + (n – 1)(-3) \implies -225 = -3(n – 1) \implies n – 1 = 75 \implies n = 76$$
Sum:
$$S_{76} = \frac{76}{2}[-5 + (-230)] = 38(-235) = -8930$$
Sum: $-8930$.
Question 3. In an AP:
(i) Given $a = 5, d = 3, a_n = 50$, find $n$ and $S_n$.
(ii) Given $a = 7, a_{13} = 35$, find $d$ and $S_{13}$.
(iii) Given $a_{12} = 37, d = 3$, find $a$ and $S_{12}$.
(iv) Given $a_3 = 15, S_{10} = 125$, find $d$ and $a_{10}$.
(v) Given $d = 5, S_9 = 75$, find $a$ and $a_9$.
(vi) Given $a = 2, d = 8, S_n = 90$, find $n$ and $a_n$.
(vii) Given $a = 8, a_n = 62, S_n = 210$, find $n$ and $d$.
(viii) Given $a_n = 4, d = 2, S_n = -14$, find $n$ and $a$.
(ix) Given $a = 3, n = 8, S = 192$, find $d$.
(x) Given $l = 28, S = 144$, and there are total $9$ terms. Find $a$. (Page No. 112-113)
Answer:
- (i) $50 = 5 + (n – 1)3 \implies 3(n – 1) = 45 \implies n = 16$.$S_{16} = \frac{16}{2}(5 + 50) = 8(55) = 440$.$n = 16, S_n = 440$.
- (ii) $35 = 7 + 12d \implies 12d = 28 \implies d = \frac{7}{3}$.$S_{13} = \frac{13}{2}(7 + 35) = \frac{13 \times 42}{2} = 273$.$d = \frac{7}{3}, S_{13} = 273$.
- (iii) $37 = a + 11(3) \implies a = 37 – 33 = 4$.$S_{12} = \frac{12}{2}(4 + 37) = 6(41) = 246$.$a = 4, S_{12} = 246$.
- (iv) $a + 2d = 15$ and $S_{10} = \frac{10}{2}[2a + 9d] = 125 \implies 2a + 9d = 25$.Multiplying first by 2: $2a + 4d = 30$. Subtracting gives $5d = -5 \implies d = -1$.Then $a = 15 – 2(-1) = 17$, and $a_{10} = 17 + 9(-1) = 8$.$d = -1, a_{10} = 8$.
- (v) $75 = \frac{9}{2}[2a + 8(5)] \implies 75 = \frac{9}{2}(2a + 40) = 9a + 180 \implies 9a = -105 \implies a = -\frac{35}{3}$.$a_9 = -\frac{35}{3} + 8(5) = \frac{85}{3}$.$a = -\frac{35}{3}, a_9 = \frac{85}{3}$.
- (vi) $90 = \frac{n}{2}[2(2) + (n – 1)8] \implies 90 = n(2 + 4n – 4) = n(4n – 2) \implies 4n^2 – 2n – 90 = 0 \implies 2n^2 – n – 45 = 0$.$(2n + 9)(n – 5) = 0 \implies n = 5$ (rejecting $n = -\frac{9}{2}$).$a_5 = 2 + 4(8) = 34$.$n = 5, a_n = 34$.
- (vii) $S_n = \frac{n}{2}(a + a_n) \implies 210 = \frac{n}{2}(8 + 62) = 35n \implies n = 6$.$62 = 8 + 5d \implies 5d = 54 \implies d = \frac{54}{5}$.$n = 6, d = \frac{54}{5}$.
- (viii) $a_n = a + (n – 1)2 = 4 \implies a = 6 – 2n$.$S_n = \frac{n}{2}(a + 4) = -14 \implies n(6 – 2n + 4) = -28 \implies n(10 – 2n) = -28 \implies 2n^2 – 10n – 28 = 0 \implies n^2 – 5n – 14 = 0$.$(n – 7)(n + 2) = 0 \implies n = 7$ (rejecting $n = -2$).$a = 6 – 2(7) = -8$.$n = 7, a = -8$.
- (ix) $192 = \frac{8}{2}[2(3) + 7d] = 4(6 + 7d) \implies 24 + 28d = 192 \implies 28d = 168 \implies d = 6$.$d = 6$.
- (x) $144 = \frac{9}{2}(a + 28) \implies a + 28 = \frac{144 \times 2}{9} = 32 \implies a = 4$.$a = 4$.
Question 4. How many terms of the AP: $9, 17, 25, \dots$ must be taken to give a sum of $636$? (Page No. 113)
Answer:
Here $a = 9, d = 8$, and $S_n = 636$.
$$\begin{aligned} 636 &= \frac{n}{2}[2(9) + (n – 1)8] \\ 636 &= n[9 + 4(n – 1)] = n(4n + 5) \\ 4n^2 + 5n – 636 &= 0 \end{aligned}$$
Factoring using the quadratic formula:
$$n = \frac{-5 \pm \sqrt{25 – 4(4)(-636)}}{8} = \frac{-5 \pm \sqrt{25 + 10176}}{8} = \frac{-5 \pm 101}{8}$$
$$n = \frac{96}{8} = 12 \quad \text{or} \quad n = -\frac{106}{8} \text{ (rejected)}$$
Thus, $n = 12$ terms must be taken.
Question 5. The first term of an AP is $5$, the last term is $45$ and the sum is $400$. Find the number of terms and the common difference. (Page No. 113)
Answer:
Given $a = 5, l = 45, S_n = 400$.
$$S_n = \frac{n}{2}(a + l) \implies 400 = \frac{n}{2}(5 + 45) = 25n \implies n = 16$$
Using $l = a + (n – 1)d$:
$$45 = 5 + 15d \implies 15d = 40 \implies d = \frac{40}{15} = \frac{8}{3}$$
Number of terms: $n = 16$; Common difference: $d = \frac{8}{3}$.
Question 6. The first and the last terms of an AP are $17$ and $350$ respectively. If the common difference is $9$, how many terms are there and what is their sum? (Page No. 113)
Answer:
Given $a = 17, l = 350, d = 9$.
$$350 = 17 + (n – 1)9 \implies 9(n – 1) = 333 \implies n – 1 = 37 \implies n = 38$$
Sum:
$$S_{38} = \frac{38}{2}(17 + 350) = 19(367) = 6973$$
Number of terms: $38$; Sum: $6973$.
Question 7. Find the sum of first $22$ terms of an AP in which $d = 7$ and $22^{\text{nd}}$ term is $149$. (Page No. 113)
Answer:
Given $n = 22, d = 7, a_{22} = 149$.
$$149 = a + 21(7) = a + 147 \implies a = 2$$
Sum:
$$S_{22} = \frac{22}{2}(a + a_{22}) = 11(2 + 149) = 11(151) = 1661$$
The sum of the first 22 terms is $1661$.
Question 8. Find the sum of first $51$ terms of an AP whose second and third terms are $14$ and $18$ respectively. (Page No. 113)
Answer:
Here $d = a_3 – a_2 = 18 – 14 = 4$.
First term: $a = a_2 – d = 14 – 4 = 10$.
$$S_{51} = \frac{51}{2}[2(10) + (51 – 1)4] = \frac{51}{2}[20 + 200] = \frac{51 \times 220}{2} = 51 \times 110 = 5610$$
The sum of the first 51 terms is $5610$.
Question 9. If the sum of first $7$ terms of an AP is $49$ and that of $17$ terms is $289$, find the sum of first $n$ terms. (Page No. 113)
Answer:
Given $S_7 = 49$ and $S_{17} = 289$.
$$\frac{7}{2}[2a + 6d] = 49 \implies 2a + 6d = 14 \implies a + 3d = 7 \quad \text{— (1)}$$
$$\frac{17}{2}[2a + 16d] = 289 \implies 2a + 16d = 34 \implies a + 8d = 17 \quad \text{— (2)}$$
Subtracting (1) from (2): $5d = 10 \implies d = 2$.
Then $a = 7 – 3(2) = 1$.
$$S_n = \frac{n}{2}[2(1) + (n – 1)2] = \frac{n}{2}[2n] = n^2$$
The sum of the first $n$ terms is $S_n = n^2$.
Question 10. Show that $a_1, a_2, \dots, a_n, \dots$ form an AP where $a_n$ is defined as below:
(i) $a_n = 3 + 4n$
(ii) $a_n = 9 – 5n$
Also find the sum of the first $15$ terms in each case. (Page No. 113)
Answer:
(i) $a_n = 3 + 4n$:
$a_{n} – a_{n-1} = (3 + 4n) – [3 + 4(n – 1)] = 4$, which is constant. Hence it forms an AP with $a = 7, d = 4$.
$$S_{15} = \frac{15}{2}[2(7) + 14(4)] = \frac{15}{2}[14 + 56] = \frac{15 \times 70}{2} = 525$$
Sum $S_{15} = 525$.
(ii) $a_n = 9 – 5n$:
$a_n – a_{n-1} = (9 – 5n) – [9 – 5(n – 1)] = -5$, which is constant. Hence it forms an AP with $a = 4, d = -5$.
$$S_{15} = \frac{15}{2}[2(4) + 14(-5)] = \frac{15}{2}[8 – 70] = \frac{15(-62)}{2} = -465$$
Sum $S_{15} = -465$.
Question 11. If the sum of the first $n$ terms of an AP is $4n – n^2$, what is the first term (that is $S_1$)? What is the sum of first two terms? What is the second term? Similarly, find the $3^{\text{rd}}$, the $10^{\text{th}}$ and the $n^{\text{th}}$ terms. (Page No. 113)
Answer:
Given $S_n = 4n – n^2$.
- $S_1 = 4(1) – 1^2 = 3 \implies a_1 = 3$.
- $S_2 = 4(2) – 2^2 = 8 – 4 = 4$.
- $a_2 = S_2 – S_1 = 4 – 3 = 1$.
- $d = a_2 – a_1 = 1 – 3 = -2$.
- $a_3 = a + 2d = 3 + 2(-2) = -1$.
- $a_{10} = a + 9d = 3 + 9(-2) = -15$.
- $a_n = S_n – S_{n-1} = (4n – n^2) – [4(n – 1) – (n – 1)^2] = 5 – 2n$.
Results: First term $3$, sum of first two terms $4$, second term $1$, third term $-1$, tenth term $-15$, $n^{\text{th}}$ term $5 – 2n$.
Question 12. Find the sum of the first $40$ positive integers divisible by $6$. (Page No. 113)
Answer:
The sequence is $6, 12, 18, \dots$ to $40$ terms. Here $a = 6, d = 6, n = 40$.
$$S_{40} = \frac{40}{2}[2(6) + 39(6)] = 20[12 + 234] = 20(246) = 4920$$
The sum is $4920$.
Question 13. Find the sum of the first $15$ multiples of $8$. (Page No. 113)
Answer:
The sequence is $8, 16, 24, \dots$ to $15$ terms. Here $a = 8, d = 8, n = 15$.
$$S_{15} = \frac{15}{2}[2(8) + 14(8)] = \frac{15}{2}[16 + 112] = \frac{15 \times 128}{2} = 15 \times 64 = 960$$
The sum is $960$.
Question 14. Find the sum of the odd numbers between $0$ and $50$. (Page No. 113)
Answer:
The odd numbers are $1, 3, 5, \dots, 49$. Here $a = 1, d = 2, l = 49$.
$$49 = 1 + (n – 1)2 \implies 2(n – 1) = 48 \implies n = 25$$
$$S_{25} = \frac{25}{2}(1 + 49) = \frac{25 \times 50}{2} = 625$$
The sum is $625$.
Question 15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ $200$ for the first day, ₹ $250$ for the second day, ₹ $300$ for the third day, etc., the penalty for each succeeding day being ₹ $50$ more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by $30$ days? (Page No. 113)
Answer:
The daily penalties form an AP: $200, 250, 300, \dots$ for $n = 30$ days.
Here $a = 200, d = 50, n = 30$.
$$S_{30} = \frac{30}{2}[2(200) + 29(50)] = 15[400 + 1450] = 15(1850) = 27750$$
The contractor has to pay ₹ $27,750$ as penalty.
Question 16. A sum of ₹ $700$ is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ $20$ less than its preceding prize, find the value of each of the prizes. (Page No. 113)
Answer:
Let the 7 prizes be $a, a – 20, a – 40, \dots$ with $n = 7, d = -20, S_7 = 700$.
$$700 = \frac{7}{2}[2a + 6(-20)] \implies 100 = \frac{1}{2}(2a – 120) \implies a – 60 = 100 \implies a = 160$$
The values of the seven prizes are:
₹ $160$, ₹ $140$, ₹ $120$, ₹ $100$, ₹ $80$, ₹ $60$, and ₹ $40$.
Question 17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant $1$ tree, a section of Class II will plant $2$ trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students? (Page No. 114)
Answer:
Each class has 3 sections.
Number of trees planted by Class $k = 3 \times k$.
Sequence: $3(1), 3(2), 3(3), \dots, 3(12) \implies 3, 6, 9, \dots, 36$.
This is an AP with $a = 3, d = 3, n = 12, l = 36$.
$$S_{12} = \frac{12}{2}(a + l) = 6(3 + 36) = 6(39) = 234$$
Total number of trees planted is $234$.
Question 18. A spiral is made up of successive semicircles, with centres alternately at $A$ and $B$, starting with centre at $A$, of radii $0.5\text{ cm}, 1.0\text{ cm}, 1.5\text{ cm}, 2.0\text{ cm}, \dots$ What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take $\pi = \frac{22}{7}$) (Page No. 114)
Answer:
Length of a semicircle of radius $r$ is $l = \pi r$.
For radii $r_1 = 0.5, r_2 = 1.0, r_3 = 1.5, \dots, r_{13} = 6.5$:
Lengths: $l_1 = 0.5\pi, l_2 = 1.0\pi, l_3 = 1.5\pi, \dots, l_{13} = 6.5\pi$.
This forms an AP with $a = 0.5\pi$, $d = 0.5\pi$, and $n = 13$.
$$S_{13} = \frac{13}{2}[2(0.5\pi) + 12(0.5\pi)] = \frac{13}{2}[\pi + 6\pi] = \frac{13}{2}(7\pi) = \frac{13}{2} \times 7 \times \frac{22}{7} = 143\text{ cm}$$
Total length of the spiral is $143\text{ cm}$.
Question 19. $200$ logs are stacked in the following manner: $20$ logs in the bottom row, $19$ in the next row, $18$ in the row next to it and so on. In how many rows are the $200$ logs placed and how many logs are in the top row? (Page No. 114)
Answer:
The number of logs per row forms an AP: $20, 19, 18, \dots$ with $a = 20, d = -1, S_n = 200$.
$$\begin{aligned} 200 &= \frac{n}{2}[2(20) + (n – 1)(-1)] \\ 400 &= n(41 – n) \\ n^2 – 41n + 400 &= 0 \\ (n – 16)(n – 25) &= 0 \implies n = 16 \text{ or } n = 25 \end{aligned}$$
Checking $n = 25$: $a_{25} = 20 + 24(-1) = -4$ (physically impossible).
Checking $n = 16$: $a_{16} = 20 + 15(-1) = 5$.
The logs are placed in $16$ rows, with $5$ logs in the top row.
Question 20. In a potato race, a bucket is placed at the starting point, which is $5\text{ m}$ from the first potato, and the other potatoes are placed $3\text{ m}$ apart in a straight line. There are ten potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run? (Page No. 114)
Answer:
For each potato, the runner travels twice the distance from the bucket:
- $1^{\text{st}}$ potato: $2 \times 5 = 10\text{ m}$
- $2^{\text{nd}}$ potato: $2 \times (5 + 3) = 16\text{ m}$
- $3^{\text{rd}}$ potato: $2 \times (5 + 6) = 22\text{ m}$
This forms an AP: $10, 16, 22, \dots$ for $n = 10$ potatoes, with $a = 10, d = 6, n = 10$.
$$S_{10} = \frac{10}{2}[2(10) + 9(6)] = 5[20 + 54] = 5(74) = 370\text{ m}$$
The total distance the competitor has to run is $370\text{ m}$.
5. Exercise 5.4 (Optional)* Solutions (Page No. 115-116)
Question 1. Which term of the AP: $121, 117, 113, \dots$, is its first negative term? (Page No. 115)
Answer:
Here $a = 121$ and $d = 117 – 121 = -4$.
We seek the smallest integer $n$ such that $a_n < 0$.
$$\begin{aligned} a + (n – 1)d &< 0 \\ 121 + (n – 1)(-4) &< 0 \\ 121 – 4n + 4 &< 0 \\ 125 &< 4n \implies n > \frac{125}{4} = 31.25 \end{aligned}$$
Since $n$ must be an integer, the smallest integer greater than $31.25$ is $n = 32$.
The $32^{\text{nd}}$ term is the first negative term.
Question 2. The sum of the third and the seventh terms of an AP is $6$ and their product is $8$. Find the sum of first sixteen terms of the AP. (Page No. 115)
Answer:
Let the terms be $a_3 = a + 2d$ and $a_7 = a + 6d$.
$$a_3 + a_7 = 6 \implies 2a + 8d = 6 \implies a + 4d = 3 \implies a = 3 – 4d$$
Their product:
$$\begin{aligned} (a + 2d)(a + 6d) &= 8 \\ (3 – 2d)(3 + 2d) &= 8 \\ 9 – 4d^2 &= 8 \implies 4d^2 = 1 \implies d = \pm \frac{1}{2} \end{aligned}$$
- Case 1: $d = \frac{1}{2} \implies a = 3 – 4\left(\frac{1}{2}\right) = 1$$$S_{16} = \frac{16}{2}\left[2(1) + 15\left(\frac{1}{2}\right)\right] = 8\left[2 + \frac{15}{2}\right] = 8\left(\frac{19}{2}\right) = 76$$
- Case 2: $d = -\frac{1}{2} \implies a = 3 – 4\left(-\frac{1}{2}\right) = 5$$$S_{16} = \frac{16}{2}\left[2(5) + 15\left(-\frac{1}{2}\right)\right] = 8\left[10 – \frac{15}{2}\right] = 8\left(\frac{5}{2}\right) = 20$$
The sum of the first 16 terms is $76$ or $20$.
Question 3. A ladder has rungs $25\text{ cm}$ apart. The rungs decrease uniformly in length from $45\text{ cm}$ at the bottom to $25\text{ cm}$ at the top. If the top and the bottom rungs are $2\frac{1}{2}\text{ m}$ apart, what is the length of the wood required for the rungs? (Page No. 115)
Answer:
Total distance between top and bottom rungs $= 2.5\text{ m} = 250\text{ cm}$.
Distance between adjacent rungs $= 25\text{ cm}$.
Number of rungs:
$$n = \frac{250}{25} + 1 = 10 + 1 = 11$$
The rung lengths form an AP with $n = 11$, first term $a = 45\text{ cm}$, and last term $l = 25\text{ cm}$.
$$S_{11} = \frac{11}{2}(a + l) = \frac{11}{2}(45 + 25) = \frac{11}{2}(70) = 385\text{ cm} = 3.85\text{ m}$$
The total length of wood required is $385\text{ cm}$ (or $3.85\text{ m}$).
Question 4. The houses of a row are numbered consecutively from $1$ to $49$. Show that there is a value of $x$ such that the sum of the numbers of the houses preceding the house numbered $x$ is equal to the sum of the numbers of the houses following it. Find this value of $x$. (Page No. 115)
Answer:
The house numbers form the sequence $1, 2, 3, \dots, 49$.
Sum of houses preceding $x$ is $S_{x-1}$.
Sum of houses following $x$ is $S_{49} – S_x$.
According to the condition:
$$\begin{aligned} S_{x-1} &= S_{49} – S_x \\ S_{x-1} + S_x &= S_{49} \\ \frac{(x – 1)x}{2} + \frac{x(x + 1)}{2} &= \frac{49 \times 50}{2} \\ \frac{x^2 – x + x^2 + x}{2} &= 1225 \\ x^2 &= 1225 \implies x = 35 \quad (x > 0) \end{aligned}$$
The value of $x$ is $35$.
Question 5. A small terrace at a football ground comprises of $15$ steps each of which is $50\text{ m}$ long and built of solid concrete. Each step has a rise of $\frac{1}{4}\text{ m}$ and a tread of $\frac{1}{2}\text{ m}$. Calculate the total volume of concrete required to build the terrace. (Page No. 116)
Answer:
For each step: length $= 50\text{ m}$, width (tread) $= \frac{1}{2}\text{ m}$.
The height increases uniformly:
- $1^{\text{st}}$ step height $= \frac{1}{4}\text{ m} \implies V_1 = 50 \times \frac{1}{2} \times \frac{1}{4} = \frac{25}{4}\text{ m}^3$
- $2^{\text{nd}}$ step height $= 2 \times \frac{1}{4}\text{ m} \implies V_2 = 2 \times \frac{25}{4}\text{ m}^3$
- $15^{\text{th}}$ step height $= 15 \times \frac{1}{4}\text{ m} \implies V_{15} = 15 \times \frac{25}{4}\text{ m}^3$
This forms an AP of volumes with $a = \frac{25}{4}, d = \frac{25}{4}, n = 15$.
$$S_{15} = \frac{15}{2}\left[\frac{25}{4} + \frac{375}{4}\right] = \frac{15}{2}\left[\frac{400}{4}\right] = \frac{15}{2}(100) = 750\text{ m}^3$$
The total volume of concrete required is $750\text{ m}^3$.
6. 15 High-Yield Frequently Asked Questions (Board Exam Level FAQs)
Question 1. If the sum of first $m$ terms of an AP is the same as the sum of its first $n$ terms ($m \neq n$), show that the sum of its first $(m + n)$ terms is zero. [CBSE Standard 2020, 2023]
Answer:
Given $S_m = S_n$.
$$\begin{aligned} \frac{m}{2}[2a + (m – 1)d] &= \frac{n}{2}[2a + (n – 1)d] \\ 2am + m(m – 1)d &= 2an + n(n – 1)d \\ 2a(m – n) + [m^2 – m – n^2 + n]d &= 0 \\ 2a(m – n) + [(m – n)(m + n) – (m – n)]d &= 0 \\ (m – n)[2a + (m + n – 1)d] &= 0 \end{aligned}$$
Since $m \neq n$, we have $m – n \neq 0$. Dividing both sides by $(m – n)$:
$$2a + (m + n – 1)d = 0$$
Now, computing the sum of the first $(m + n)$ terms:
$$S_{m+n} = \frac{m + n}{2}[2a + (m + n – 1)d] = \frac{m + n}{2}(0) = 0$$
Hence proved.
Question 2. The ratio of the $11^{\text{th}}$ term to the $18^{\text{th}}$ term of an AP is $2 : 3$. Find the ratio of the $5^{\text{th}}$ term to the $21^{\text{st}}$ term and the ratio of the sum of the first $5$ terms to the sum of the first $21$ terms. [CBSE 2019]
Answer:
Given $\frac{a_{11}}{a_{18}} = \frac{2}{3} \implies \frac{a + 10d}{a + 17d} = \frac{2}{3}$.
$$3a + 30d = 2a + 34d \implies a = 4d$$
- Ratio of $5^{\text{th}}$ to $21^{\text{st}}$ term:$$\frac{a_5}{a_{21}} = \frac{a + 4d}{a + 20d} = \frac{4d + 4d}{4d + 20d} = \frac{8d}{24d} = \frac{1}{3}$$
- Ratio of sums $S_5$ to $S_{21}$:$$\frac{S_5}{S_{21}} = \frac{\frac{5}{2}[2(4d) + 4d]}{\frac{21}{2}[2(4d) + 20d]} = \frac{5(12d)}{21(28d)} = \frac{60}{588} = \frac{5}{49}$$Ratios are $1 : 3$ and $5 : 49$.
Question 3. If the $p^{\text{th}}$ term of an AP is $\frac{1}{q}$ and the $q^{\text{th}}$ term is $\frac{1}{p}$, prove that the sum of first $pq$ terms is $\frac{1}{2}(pq + 1)$, where $p \neq q$. [CBSE 2017, 2020]
Answer:
Given $a_p = a + (p – 1)d = \frac{1}{q}$ and $a_q = a + (q – 1)d = \frac{1}{p}$.
Subtracting the two equations:
$$(p – q)d = \frac{1}{q} – \frac{1}{p} = \frac{p – q}{pq} \implies d = \frac{1}{pq}$$
Substituting $d$ back:
$$a = \frac{1}{q} – (p – 1)\frac{1}{pq} = \frac{p – p + 1}{pq} = \frac{1}{pq}$$
Now calculating $S_{pq}$:
$$S_{pq} = \frac{pq}{2}\left[2\left(\frac{1}{pq}\right) + (pq – 1)\left(\frac{1}{pq}\right)\right] = \frac{pq}{2}\left[\frac{2 + pq – 1}{pq}\right] = \frac{pq + 1}{2}$$
Hence proved.
Question 4. The sum of four consecutive numbers in an AP is $32$ and the ratio of the product of the first and the last term to the product of two middle terms is $7 : 15$. Find the numbers. [CBSE 2018, 2020]
Answer:
Let the four consecutive terms be $(a – 3d), (a – d), (a + d), (a + 3d)$.
Sum $= 4a = 32 \implies a = 8$.
Given ratio:
$$\begin{aligned} \frac{(a – 3d)(a + 3d)}{(a – d)(a + d)} &= \frac{7}{15} \\ \frac{a^2 – 9d^2}{a^2 – d^2} &= \frac{7}{15} \\ 15(64 – 9d^2) &= 7(64 – d^2) \\ 960 – 135d^2 &= 448 – 7d^2 \\ 128d^2 &= 512 \implies d^2 = 4 \implies d = \pm 2 \end{aligned}$$
For $a = 8, d = 2$, the numbers are: $8 – 6, 8 – 2, 8 + 2, 8 + 6 \implies 2, 6, 10, 14$.
The numbers are $2, 6, 10, 14$.
Question 5. If $S_n$ denotes the sum of the first $n$ terms of an AP, prove that $S_{12} = 3(S_8 – S_4)$. [CBSE 2015, 2019]
Answer:
Using $S_n = \frac{n}{2}[2a + (n – 1)d]$:
$$S_8 = \frac{8}{2}[2a + 7d] = 4(2a + 7d) = 8a + 28d$$
$$S_4 = \frac{4}{2}[2a + 3d] = 2(2a + 3d) = 4a + 6d$$
Subtracting $S_4$ from $S_8$:
$$S_8 – S_4 = (8a + 28d) – (4a + 6d) = 4a + 22d$$
Multiplying by $3$:
$$3(S_8 – S_4) = 3(4a + 22d) = 12a + 66d$$
Now computing $S_{12}$:
$$S_{12} = \frac{12}{2}[2a + 11d] = 6(2a + 11d) = 12a + 66d$$
Thus, $S_{12} = 3(S_8 – S_4)$. Hence proved.
Question 6. The ratio of the sums of first $m$ and first $n$ terms of an AP is $m^2 : n^2$. Show that the ratio of its $m^{\text{th}}$ and $n^{\text{th}}$ terms is $(2m – 1) : (2n – 1)$. [CBSE 2016, 2017, 2023]
Answer:
Given $\frac{S_m}{S_n} = \frac{m^2}{n^2} \implies \frac{\frac{m}{2}[2a + (m – 1)d]}{\frac{n}{2}[2a + (n – 1)d]} = \frac{m^2}{n^2} \implies \frac{2a + (m – 1)d}{2a + (n – 1)d} = \frac{m}{n}$.
Cross-multiplying:
$$\begin{aligned} n[2a + (m – 1)d] &= m[2a + (n – 1)d] \\ 2an + nmd – nd &= 2am + mnd – md \\ 2a(n – m) &= d(n – m) \implies d = 2a \end{aligned}$$
Ratio of $m^{\text{th}}$ to $n^{\text{th}}$ term:
$$\frac{a_m}{a_n} = \frac{a + (m – 1)d}{a + (n – 1)d} = \frac{a + (m – 1)(2a)}{a + (n – 1)(2a)} = \frac{a(1 + 2m – 2)}{a(1 + 2n – 2)} = \frac{2m – 1}{2n – 1}$$
Hence proved.
Question 7. Find the sum of all two-digit natural numbers which leave a remainder $2$ when divided by $3$.
Answer:
The smallest two-digit number satisfying the condition is $11$, and the largest is $98$.
Sequence: $11, 14, 17, \dots, 98$.
Here $a = 11, d = 3, l = 98$.
$$98 = 11 + (n – 1)3 \implies 3(n – 1) = 87 \implies n = 30$$
$$S_{30} = \frac{30}{2}(11 + 98) = 15(109) = 1635$$
The sum is $1635$.
Question 8. [Assertion-Reason]
Assertion (A): The $n^{\text{th}}$ term of the AP $5, 2, -1, -4, \dots$ is $8 – 3n$.
Reason (R): The $n^{\text{th}}$ term of an AP with first term $a$ and common difference $d$ is given by $a_n = a + (n – 1)d$.
Answer:
- Evaluating Assertion: Here $a = 5, d = 2 – 5 = -3$.$a_n = 5 + (n – 1)(-3) = 5 – 3n + 3 = 8 – 3n$. Assertion (A) is true.
- Evaluating Reason: The formula $a_n = a + (n – 1)d$ is correct and directly explains (A).Correct Choice: Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Question 9. [Assertion-Reason]
Assertion (A): The sum of the first $n$ natural numbers is given by $\frac{n(n + 1)}{2}$.
Reason (R): If $a, b, c$ are in AP, then $b = \frac{a + c}{2}$.
Answer:
- Evaluating Assertion: Natural numbers form an AP $1, 2, 3, \dots, n$ with $a = 1, d = 1$. $S_n = \frac{n}{2}(1 + n) = \frac{n(n + 1)}{2}$. (True)
- Evaluating Reason: If $a, b, c$ are in AP, $b – a = c – b \implies 2b = a + c \implies b = \frac{a + c}{2}$. (True, but not the direct mathematical proof of the summation formula).Correct Choice: Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
Question 10. [Case Study 1] A manufacturer of laptop batteries produces $600$ units in the third year and $700$ units in the seventh year. Assuming that the production increases uniformly by a constant number every year, find:
(i) The production in the $1^{\text{st}}$ year.
(ii) The production in the $10^{\text{th}}$ year.
(iii) The total production in first $7$ years.
Answer:
Given $a_3 = a + 2d = 600$ and $a_7 = a + 6d = 700$.
Subtracting gives $4d = 100 \implies d = 25$.
- (i) $a = 600 – 2(25) = 550$ units. Production in $1^{\text{st}}$ year is $550\text{ units}$.
- (ii) $a_{10} = 550 + 9(25) = 550 + 225 = 775$ units. Production in $10^{\text{th}}$ year is $775\text{ units}$.
- (iii) $S_7 = \frac{7}{2}(a + a_7) = \frac{7}{2}(550 + 700) = \frac{7 \times 1250}{2} = 4375$ units. Total production in first 7 years is $4375\text{ units}$.
Question 11. Solve the equation for $x$: $1 + 4 + 7 + 10 + \dots + x = 287$. [CBSE 2020]
Answer:
Here $a = 1, d = 3$, and $S_n = 287$, where last term $x = a_n = 1 + (n – 1)3 = 3n – 2$.
$$\begin{aligned} S_n &= \frac{n}{2}(a + l) = \frac{n}{2}(1 + 3n – 2) = 287 \\ n(3n – 1) &= 574 \\ 3n^2 – n – 574 &= 0 \end{aligned}$$
Factoring:
$$n = \frac{1 \pm \sqrt{1 – 4(3)(-574)}}{6} = \frac{1 \pm \sqrt{1 + 6888}}{6} = \frac{1 \pm 83}{6}$$
$$n = \frac{84}{6} = 14 \quad (\text{rejecting } n = -\frac{82}{6})$$
Then $x = a_{14} = 1 + 13(3) = 1 + 39 = 40$.
The value of $x$ is $40$.
Question 12. An AP consists of $37$ terms. The sum of the three middle-most terms is $225$ and the sum of the last three terms is $429$. Find the AP. [CBSE 2017]
Answer:
With $n = 37$, the middle term is $\frac{37 + 1}{2} = 19^{\text{th}}$ term.
The three middle-most terms are $a_{18}, a_{19}, a_{20}$.
$$a_{18} + a_{19} + a_{20} = 3a_{19} = 225 \implies a_{19} = a + 18d = 75 \quad \text{— (1)}$$
The last three terms are $a_{35}, a_{36}, a_{37}$.
$$(a + 34d) + (a + 35d) + (a + 36d) = 3a + 105d = 429 \implies a + 35d = 143 \quad \text{— (2)}$$
Subtracting (1) from (2):
$$17d = 68 \implies d = 4$$
Substituting $d = 4$ into (1): $a = 75 – 18(4) = 75 – 72 = 3$.
The AP is $3, 7, 11, 15, \dots$
Question 13. Divide $56$ into four parts in AP such that the ratio of the product of their extremes ($1^{\text{st}}$ and $4^{\text{th}}$) to the product of their means ($2^{\text{nd}}$ and $3^{\text{rd}}$) is $5 : 6$. [CBSE 2016]
Answer:
Let the four parts be $(a – 3d), (a – d), (a + d), (a + 3d)$.
Sum $= 4a = 56 \implies a = 14$.
Ratio condition:
$$\begin{aligned} \frac{(14 – 3d)(14 + 3d)}{(14 – d)(14 + d)} &= \frac{5}{6} \\ \frac{196 – 9d^2}{196 – d^2} &= \frac{5}{6} \\ 1176 – 54d^2 &= 980 – 5d^2 \\ 49d^2 &= 196 \implies d^2 = 4 \implies d = \pm 2 \end{aligned}$$
For $a = 14, d = 2$: the parts are $14 – 6, 14 – 2, 14 + 2, 14 + 6 \implies 8, 12, 16, 20$.
The four parts are $8, 12, 16, 20$.
Question 14. Find the sum of all natural numbers between $200$ and $400$ which are divisible by $7$.
Answer:
The numbers strictly between $200$ and $400$ divisible by $7$ are $203, 210, 217, \dots, 399$.
Here $a = 203, d = 7, l = 399$.
$$399 = 203 + (n – 1)7 \implies 7(n – 1) = 196 \implies n – 1 = 28 \implies n = 29$$
$$S_{29} = \frac{29}{2}(203 + 399) = \frac{29 \times 602}{2} = 29 \times 301 = 8729$$
The sum is $8729$.
Question 15. [Case Study 2] In an auditorium, seats are arranged in rows such that the first row has $20$ seats, the second row has $24$ seats, the third row has $28$ seats, and so on. If there are $30$ rows in total:
(i) Find the number of seats in the $15^{\text{th}}$ row.
(ii) Find the total seating capacity of the auditorium.
(iii) If 1500 people want to attend a show, how many more rows of seats are needed?
Answer:
Here $a = 20, d = 4, n = 30$.
- (i) $a_{15} = 20 + 14(4) = 20 + 56 = 76$. There are $76$ seats in the $15^{\text{th}}$ row.
- (ii) Total capacity $S_{30}$:$$S_{30} = \frac{30}{2}[2(20) + 29(4)] = 15[40 + 116] = 15(156) = 2340\text{ seats}$$Total capacity is $2340\text{ seats}$.
- (iii) Since the current capacity ($2340$) is already greater than $1500$, all $1500$ attendees can be seated within the existing arrangement without needing additional rows. No extra rows are needed.
7. Concluding Board Topper Strategy
To attain a perfect score in Arithmetic Progressions on the CBSE Class 10 Board Exam:
- Explicit Parameter Declaration: Always begin by writing
Given: a = ..., d = ..., n = ...and defining your variables clearly. - Formula Quotation: State the standard formula ($a_n = a + (n – 1)d$ or $S_n = \frac{n}{2}[2a + (n – 1)d]$) before substituting values to secure step marks.
- Validity Checks on $n$: Always verify that $n$ is a positive integer ($n \in \mathbb{N}$). If a quadratic equation yields negative or fractional roots for $n$, explicitly write: “Rejecting $n = \dots$ as the number of terms must be a natural number.”
- Symmetric Variable Selection: When given problems involving sums of 3 or 4 consecutive terms, use $(a – d), a, (a + d)$ or $(a – 3d), (a – d), (a + d), (a + 3d)$ to eliminate $d$ immediately upon summation.
- Final Unit Boxing: Box your final numeric answer along with the correct unit ($\text{cm}$, $\text{m}^3$, $\text{₹}$, or $\text{terms}$) to keep your presentation clean and examiner-friendly.