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H1 Title: NCERT Solutions for Class 9 Math Chapter 8: Predicting What Comes Next: Exploring Sequences and Progressions (Complete Guide)
Navigating through the CBSE Class 9 Mathematics curriculum requires an in-depth understanding of inductive pattern recognition, recursive relations, explicit general formulas, and arithmetic and geometric progressions. Chapter 8 of Class 9 Mathematics, “Predicting What Comes Next: Exploring Sequences and Progressions”, forms the foundation of discrete mathematics, computer programming algorithms, financial interest calculations, and higher-level series analysis. It investigates the transition from visual dot arrangements to formal algebraic sequences; explores the mathematical behavior of constant-difference arithmetic progressions and multiplying-factor geometric sequences; and details the quantitative determination of the $n$-th term and partial sums of series. To help students master every aspect of this high-weightage chapter, this comprehensive guide offers textbook-accurate, highly structured, and pedagogically sound responses strictly aligned with the latest CBSE evaluation standards.
Every question presented in the official NCERT textbook—ranging from sequence classification and visual dot models to multi-step word problems on auditoriums and savings schemes, along with an expanded set of 15 board-level FAQs—has been solved with exhaustive detail. Key scoring terms, systematic four-step mathematical workflows (Given Data $\rightarrow$ Formula Stated $\rightarrow$ Step-by-Step LaTeX Substitution $\rightarrow$ Final Answer with Units), and clear inline geometric diagrams have been highlighted to ensure students secure maximum marks in their CBSE examinations.
Chapter 8: Predicting What Comes Next: Exploring Sequences and Progressions
Master Chapter Summary & Quantitative Blueprint
In the modern NEP-aligned Class 9 curriculum, Chapter 8 bridges visual patterns with algebraic formalism. Students learn how to analyze ordered lists of numbers, determine their underlying generative rules, and forecast terms arbitrarily far into the future without writing out all intermediate values.
| Progression / Sequence Type | Defining Mathematical Characteristic | General Term ($n$-th term, $a_n$) | Sum of First $n$ Terms ($S_n$) | Typical Student Pitfall | CBSE Marks Weightage |
|---|---|---|---|---|---|
| General Sequence | An ordered list of numbers following a specific rule | $a_n = f(n)$ | $\sum_{i=1}^n a_i$ | Confusing term index $n$ with term value $a_n$ | 1 to 2 Marks |
| Arithmetic Progression (AP) | Successive terms differ by a constant difference $d$ | $a_n = a + (n – 1)d$ | $S_n = \frac{n}{2}[2a + (n – 1)d] = \frac{n}{2}(a + l)$ | Using $n$ instead of $(n – 1)$ in formula | 3 to 4 Marks |
| Geometric Sequence (GP) | Successive terms have a constant ratio $r$ | $a_n = a \cdot r^{n-1}$ | $S_n = \frac{a(r^n – 1)}{r – 1} \quad (r \neq 1)$ | Treating common ratio as a common difference | 2 to 3 Marks |
| Triangular Numbers | Dot patterns forming equilateral triangles | $T_n = \frac{n(n + 1)}{2}$ | Sum of consecutive triangular numbers | Forgetting to divide by 2 | 2 to 3 Marks |
| Fibonacci Sequence | Each term is the sum of the two preceding terms | $F_n = F_{n-1} + F_{n-2} \quad (n \ge 3)$ | Converges to Golden Ratio $\phi \approx 1.618$ | Misidentifying starting seeds $F_1 = 1, F_2 = 1$ | 2 to 3 Marks |
🧠 Examiner’s Secret: In any Arithmetic Progression, always remember that $n$ (the number of terms or term position) must strictly be a positive integer ($n \in \mathbb{N} = {1, 2, 3, \dots}$). If your calculation for $n$ results in a negative number, zero, or a non-integer fraction (such as $n = \frac{53}{4}$), it indicates that the given number is not a term of the AP or an arithmetic error occurred.
Foundational Geometric Models and Sequence Architecture
Sequences and the Concept of General Term ($a_n$)
A sequence is an ordered succession of numbers arranged according to a definite rule, where each number is termed an element or term of the sequence.
The terms of a sequence are conventionally denoted by:
$$a_1, a_2, a_3, \dots, a_n, \dots$$
where $a_1$ represents the first term, $a_2$ the second term, and $a_n$ the general term (or $n$-th term) expressed as a function of the term index $n$.
Triangular Numbers Model
Triangular numbers are a sequence of figurate numbers that can be visually represented as a symmetric triangular grid of dots. T₁ = 1 T₂ = 3 T₃ = 6 T₄ = 10
Notice the recursive addition pattern:
- $T_1 = 1$
- $T_2 = 1 + 2 = 3$
- $T_3 = 1 + 2 + 3 = 6$
- $T_4 = 1 + 2 + 3 + 4 = 10$
In general, the $n$-th triangular number represents the sum of the first $n$ positive natural numbers:
$$T_n = 1 + 2 + 3 + \dots + n = \frac{n(n + 1)}{2}$$
Linear Growth vs. Exponential Growth
Linear growth adds a constant amount in each step (an Arithmetic Progression), while exponential growth multiplies by a constant factor in each step (a Geometric Progression). Linear (AP: +d) Exponential (GP: ×r) n (steps) aₙ
In an Arithmetic Progression, the graph of $a_n$ against $n$ forms a set of collinear points lying on a straight line with slope equal to the common difference $d$. In a Geometric Progression with common ratio $r > 1$, the terms accelerate upward along an exponential curve.
💡 Did You Know?: The famous “wheat and chessboard problem” tells the story of an inventor who requested 1 grain of wheat on the first square of a chessboard, 2 on the second, 4 on the third, doubling each time. This geometric sequence ($1, 2, 4, 8, \dots, 2^{63}$) sums to $2^{64} – 1 = 18,446,744,073,709,551,615$ grains of wheat—more than the entire world’s harvest for centuries!
[👉 Also Read: Class 9 Math Chapter 7 The Mathematics of Maybe: Introduction to Probability NCERT Solutions]
Step-by-Step Solutions: Core Textbook Exercises and Applied Problems
Question 1. [Generating Terms from Explicit Formulas]
Write the first four terms of the sequences whose $n$-th terms are given by:
(i) $a_n = 2n + 5$
(ii) $a_n = \frac{n – 3}{4}$
(iii) $a_n = (-1)^{n-1} 5^{n+1}$
Answer:
(i) For $a_n = 2n + 5$:
- Substitute $n = 1$: $a_1 = 2(1) + 5 = 2 + 5 = 7$
- Substitute $n = 2$: $a_2 = 2(2) + 5 = 4 + 5 = 9$
- Substitute $n = 3$: $a_3 = 2(3) + 5 = 6 + 5 = 11$
- Substitute $n = 4$: $a_4 = 2(4) + 5 = 8 + 5 = 13$
- Final Answer: The first four terms are $7, 9, 11, 13$.
(ii) For $a_n = \frac{n – 3}{4}$:
- Substitute $n = 1$: $a_1 = \frac{1 – 3}{4} = \frac{-2}{4} = -\frac{1}{2}$
- Substitute $n = 2$: $a_2 = \frac{2 – 3}{4} = -\frac{1}{4}$
- Substitute $n = 3$: $a_3 = \frac{3 – 3}{4} = \frac{0}{4} = 0$
- Substitute $n = 4$: $a_4 = \frac{4 – 3}{4} = \frac{1}{4}$
- Final Answer: The first four terms are $-\frac{1}{2}, -\frac{1}{4}, 0, \frac{1}{4}$.
(iii) For $a_n = (-1)^{n-1} 5^{n+1}$:
- Substitute $n = 1$: $a_1 = (-1)^{1-1} 5^{1+1} = (-1)^0 5^2 = (1)(25) = 25$
- Substitute $n = 2$: $a_2 = (-1)^{2-1} 5^{2+1} = (-1)^1 5^3 = (-1)(125) = -125$
- Substitute $n = 3$: $a_3 = (-1)^{3-1} 5^{3+1} = (-1)^2 5^4 = (1)(625) = 625$
- Substitute $n = 4$: $a_4 = (-1)^{4-1} 5^{4+1} = (-1)^3 5^5 = (-1)(3125) = -3125$
- Final Answer: The first four terms are $25, -125, 625, -3125$.
Question 2. [Identifying Arithmetic Progressions]
Which of the following sequences form an Arithmetic Progression (AP)? If they form an AP, find the common difference $d$ and write the next three 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) $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots$
Answer:
Testing Condition for an AP:
A sequence forms an AP if and only if the difference between consecutive terms remains constant: $a_{k+1} – a_k = d$ for all $k$.
(i) For $2, 4, 8, 16, \dots$:
- $a_2 – a_1 = 4 – 2 = 2$
- $a_3 – a_2 = 8 – 4 = 4$
- Since $a_2 – a_1 \neq a_3 – a_2$ ($2 \neq 4$), the difference is not constant.
- Conclusion: It is not an AP (it is a Geometric Progression with $r = 2$).
(ii) For $2, \frac{5}{2}, 3, \frac{7}{2}, \dots$:
- $a_2 – a_1 = \frac{5}{2} – 2 = \frac{1}{2}$
- $a_3 – a_2 = 3 – \frac{5}{2} = \frac{1}{2}$
- $a_4 – a_3 = \frac{7}{2} – 3 = \frac{1}{2}$
- Since $a_{k+1} – a_k = \frac{1}{2}$ is constant, it forms an AP with $d = \frac{1}{2}$.
- Next Three Terms:
$$a_5 = \frac{7}{2} + \frac{1}{2} = \frac{8}{2} = 4$$
$$a_6 = 4 + \frac{1}{2} = \frac{9}{2}$$
$$a_7 = \frac{9}{2} + \frac{1}{2} = 5$$ - Final Answer: It is an AP with $d = \frac{1}{2}$; next terms are $4, \frac{9}{2}, 5$.
(iii) For $-1.2, -3.2, -5.2, -7.2, \dots$:
- $a_2 – a_1 = -3.2 – (-1.2) = -3.2 + 1.2 = -2.0$
- $a_3 – a_2 = -5.2 – (-3.2) = -5.2 + 3.2 = -2.0$
- $a_4 – a_3 = -7.2 – (-5.2) = -7.2 + 5.2 = -2.0$
- Since the common difference is constant ($d = -2$), it forms an AP.
- Next Three Terms:
$$a_5 = -7.2 + (-2) = -9.2$$
$$a_6 = -9.2 + (-2) = -11.2$$
$$a_7 = -11.2 + (-2) = -13.2$$ - Final Answer: It is an AP with $d = -2$; next terms are $-9.2, -11.2, -13.2$.
(iv) For $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots$:
- Simplify radicals into like terms:
$$\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$$
$$\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$$
$$\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$$ - The sequence is $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \dots$
- $a_2 – a_1 = 2\sqrt{2} – \sqrt{2} = \sqrt{2}$
- $a_3 – a_2 = 3\sqrt{2} – 2\sqrt{2} = \sqrt{2}$
- $a_4 – a_3 = 4\sqrt{2} – 3\sqrt{2} = \sqrt{2}$
- The sequence forms an AP with common difference $d = \sqrt{2}$.
- Next Three Terms:
$$a_5 = 4\sqrt{2} + \sqrt{2} = 5\sqrt{2} = \sqrt{5^2 \times 2} = \sqrt{50}$$
$$a_6 = 5\sqrt{2} + \sqrt{2} = 6\sqrt{2} = \sqrt{6^2 \times 2} = \sqrt{72}$$
$$a_7 = 6\sqrt{2} + \sqrt{2} = 7\sqrt{2} = \sqrt{7^2 \times 2} = \sqrt{98}$$ - Final Answer: It is an AP with $d = \sqrt{2}$; next terms are $\sqrt{50}, \sqrt{72}, \sqrt{98}$.
Question 3. [Triangular Numbers Pattern Exploration]
The triangular numbers form the sequence $1, 3, 6, 10, 15, \dots$
(i) Write the recursive relationship connecting the $n$-th triangular number $T_n$ to the previous term $T_{n-1}$.
(ii) Use the explicit formula $T_n = \frac{n(n+1)}{2}$ to calculate $T_{10}$ and $T_{100}$.
(iii) Prove algebraically that the sum of two consecutive triangular numbers is always a perfect square: $T_{n-1} + T_n = n^2$.
Answer:
Part (i): Recursive Relationship
Examining the differences:
- $T_2 – T_1 = 3 – 1 = 2$
- $T_3 – T_2 = 6 – 3 = 3$
- $T_4 – T_3 = 10 – 6 = 4$
Thus, the $n$-th term is formed by adding $n$ to the previous term:
$$T_n = T_{n-1} + n \quad (\text{for } n \ge 2, \text{ with } T_1 = 1)$$
Part (ii): Numerical Calculation
- For $T_{10}$:
$$T_{10} = \frac{10(10 + 1)}{2} = \frac{10 \times 11}{2} = 5 \times 11 = 55$$ - For $T_{100}$:
$$T_{100} = \frac{100(100 + 1)}{2} = \frac{100 \times 101}{2} = 50 \times 101 = 5050$$
Part (iii): Algebraic Proof for $T_{n-1} + T_n = n^2$
- Express both terms using the explicit formula:
$$T_n = \frac{n(n + 1)}{2}$$
$$T_{n-1} = \frac{(n – 1)[(n – 1) + 1]}{2} = \frac{(n – 1)n}{2}$$ - Sum the two terms:
$$T_{n-1} + T_n = \frac{n(n – 1)}{2} + \frac{n(n + 1)}{2}$$
$$T_{n-1} + T_n = \frac{n(n – 1 + n + 1)}{2} = \frac{n(2n)}{2} = \frac{2n^2}{2} = n^2$$
Conclusion:
Hence proved, the sum of two consecutive triangular numbers is strictly equal to $n^2$.
Final Answer:
(i) Recursive rule: $T_n = T_{n-1} + n$.
(ii) $T_{10} = 55$ and $T_{100} = 5050$.
(iii) Algebraically proved that $T_{n-1} + T_n = n^2$.
Question 4. [Finding Specific Terms of an AP]
(i) Find the $10\text{-th term}$ of the AP: $2, 7, 12, \dots$
(ii) Find the $18\text{-th term}$ of the AP: $\sqrt{2}, 3\sqrt{2}, 5\sqrt{2}, \dots$
Answer:
General Formula:
$$a_n = a + (n – 1)d$$
(i) For the AP $2, 7, 12, \dots$:
- First term, $a = 2$
- Common difference, $d = 7 – 2 = 5$
- Term index, $n = 10$
- Substitution:
$$a_{10} = 2 + (10 – 1)(5) = 2 + (9)(5) = 2 + 45 = 47$$ - Final Answer: The 10-th term is $47$.
(ii) For the AP $\sqrt{2}, 3\sqrt{2}, 5\sqrt{2}, \dots$:
- First term, $a = \sqrt{2}$
- Common difference, $d = 3\sqrt{2} – \sqrt{2} = 2\sqrt{2}$
- Term index, $n = 18$
- Substitution:
$$a_{18} = \sqrt{2} + (18 – 1)(2\sqrt{2}) = \sqrt{2} + 17(2\sqrt{2}) = \sqrt{2} + 34\sqrt{2} = 35\sqrt{2}$$ - Final Answer: The 18-th term is $35\sqrt{2}$.
Question 5. [Determining Term Index $n$]
Which term of the AP: $3, 8, 13, 18, \dots$ is $78$?
Answer:
Step 1: Identify Given Parameters
- First term, $a = 3$
- Common difference, $d = 8 – 3 = 5$
- $n$-th term, $a_n = 78$
Step 2: Apply General Term Formula
$$a_n = a + (n – 1)d$$
$$78 = 3 + (n – 1)(5)$$
Step 3: Solve for $n$
Subtract $3$ from both sides:
$$78 – 3 = (n – 1)(5)$$
$$75 = 5(n – 1)$$
Divide both sides by $5$:
$$n – 1 = \frac{75}{5} = 15$$
$$n = 15 + 1 = 16$$
Since $n = 16$ is a positive integer, $78$ is the 16th term.
Final Answer:
$78$ is the $16\text{-th term}$ of the given AP.
Question 6. [Simultaneous Equations in an AP]
An AP consists of $50\text{ terms}$ of which $3\text{rd term}$ is $12$ and the last term is $106$. Find the $29\text{th term}$.
Answer:
Step 1: Identify Given Data
- Total number of terms, $n = 50$
- Third term: $a_3 = 12$
- Last term (50th term): $a_{50} = 106$
Step 2: Set Up Simultaneous Linear Equations
Using $a_n = a + (n – 1)d$:
$$a_3 = a + 2d = 12 \quad \text{— (Equation 1)}$$
$$a_{50} = a + 49d = 106 \quad \text{— (Equation 2)}$$
Step 3: Solve for $a$ and $d$
Subtract Equation 1 from Equation 2:
$$(a + 49d) – (a + 2d) = 106 – 12$$
$$47d = 94$$
$$d = \frac{94}{47} = 2$$
Substitute $d = 2$ into Equation 1:
$$a + 2(2) = 12$$
$$a + 4 = 12 \implies a = 8$$
Step 4: Compute the 29th Term ($a_{29}$)
$$a_{29} = a + (29 – 1)d = a + 28d$$
$$a_{29} = 8 + 28(2) = 8 + 56 = 64$$
Final Answer:
The 29th term of the AP is $64$.
Question 7. [Geometric Progression — Bacterial Population Growth]
In a laboratory experiment, a bacterial culture begins with an initial population of $100\text{ bacteria}$. Under optimal nutrient conditions, the population doubles every hour.
(i) Write the population values for the first four hours as a sequence.
(ii) State the general formula for the population $P_n$ after $n\text{ hours}$.
(iii) Calculate the bacterial population after $5\text{ hours}$ and after $10\text{ hours}$.
Answer:
Step 1: (i) Write the Sequence of Population
- Initial count ($t = 0$): $100$
- After $1\text{ hour}$ ($t = 1$): $100 \times 2 = 200$
- After $2\text{ hours}$ ($t = 2$): $200 \times 2 = 400$
- After $3\text{ hours}$ ($t = 3$): $400 \times 2 = 800$
- After $4\text{ hours}$ ($t = 4$): $800 \times 2 = 1600$
The sequence of hourly populations is: $200, 400, 800, 1600, \dots$
Step 2: (ii) General Formula for Population After $n$ Hours
This is a Geometric Progression where the initial value is $P_0 = 100$ and the common multiplying ratio is $r = 2$:
$$P_n = 100 \times 2^n$$
Step 3: (iii) Calculate Population at $n = 5$ and $n = 10$
- After 5 hours ($n = 5$):
$$P_5 = 100 \times 2^5 = 100 \times 32 = 3200\text{ bacteria}$$ - After 10 hours ($n = 10$):
$$P_{10} = 100 \times 2^{10} = 100 \times 1024 = 102400\text{ bacteria}$$
Final Answer:
(i) First four hours: $200, 400, 800, 1600$.
(ii) Formula: $P_n = 100 \times 2^n$.
(iii) Population after $5\text{ hours}$ is $3200$; after $10\text{ hours}$ is $102400$.
Question 8. [Fibonacci Sequence and Golden Ratio Approximations]
The Fibonacci sequence is defined by the initial seed values $F_1 = 1, F_2 = 1$ and the recurrence relation $F_n = F_{n-1} + F_{n-2}$ for $n \ge 3$.
(i) Write down the first $8\text{ terms}$ of the Fibonacci sequence.
(ii) Compute the ratio of successive terms $\frac{F_{n+1}}{F_n}$ for $n = 1, 2, 3, 4, 5$ as fractions and decimals.
Answer:
Step 1: (i) Generate First 8 Terms
- $F_1 = 1$
- $F_2 = 1$
- $F_3 = F_2 + F_1 = 1 + 1 = 2$
- $F_4 = F_3 + F_2 = 2 + 1 = 3$
- $F_5 = F_4 + F_3 = 3 + 2 = 5$
- $F_6 = F_5 + F_4 = 5 + 3 = 8$
- $F_7 = F_6 + F_5 = 8 + 5 = 13$
- $F_8 = F_7 + F_6 = 13 + 8 = 21$
The first eight terms are: $1, 1, 2, 3, 5, 8, 13, 21$.
Step 2: (ii) Calculate Successive Ratios $\frac{F_{n+1}}{F_n}$
- For $n = 1$: $\frac{F_2}{F_1} = \frac{1}{1} = 1.0$
- For $n = 2$: $\frac{F_3}{F_2} = \frac{2}{1} = 2.0$
- For $n = 3$: $\frac{F_4}{F_3} = \frac{3}{2} = 1.5$
- For $n = 4$: $\frac{F_5}{F_4} = \frac{5}{3} \approx 1.667$
- For $n = 5$: $\frac{F_6}{F_5} = \frac{8}{5} = 1.6$
(Notice how the ratios oscillate and approach the Golden Ratio $\phi \approx 1.618$).
Final Answer:
(i) First 8 terms: $1, 1, 2, 3, 5, 8, 13, 21$.
(ii) Ratios: $1, 2, 1.5, 1.667, 1.6$.
Question 9. [Sum of Natural Numbers & Arithmetic Series Formula]
(i) Find the sum of the first $100\text{ positive integers}$.
(ii) Find the sum of the first $22\text{ terms}$ of the AP: $8, 3, -2, \dots$
Answer:
(i) Sum of First 100 Positive Integers:
- $a = 1, l = 100, n = 100$
- Using Gauss’s sum formula:
$$S_n = \frac{n(n + 1)}{2}$$
$$S_{100} = \frac{100(100 + 1)}{2} = 50 \times 101 = 5050$$ - Final Answer: The sum is $5050$.
(ii) Sum of First 22 Terms of AP $8, 3, -2, \dots$:
- First term, $a = 8$
- Common difference, $d = 3 – 8 = -5$
- Number of terms, $n = 22$
- Formula:
$$S_n = \frac{n}{2}[2a + (n – 1)d]$$ - Step-by-Step Substitution:
$$S_{22} = \frac{22}{2}[2(8) + (22 – 1)(-5)]$$
$$S_{22} = 11[16 + (21)(-5)]$$
$$S_{22} = 11[16 – 105] = 11[-89] = -979$$ - Final Answer: The sum of the first 22 terms is $-979$.
Question 10. [Auditorium Seating Arrangement]
In an auditorium, there are $20\text{ seats}$ in the first row, $22\text{ seats}$ in the second row, $24\text{ seats}$ in the third row, and so on. If there are $30\text{ rows}$ of seats in total:
(i) How many seats are in the last ($30\text{th}$) row?
(ii) What is the total seating capacity of the auditorium?
Answer:
Step 1: Formulate the Arithmetic Progression
The row capacities form an AP: $20, 22, 24, \dots$
- First term, $a = 20$
- Common difference, $d = 22 – 20 = 2$
- Total rows, $n = 30$
Step 2: (i) Seats in the 30th Row ($a_{30}$)
$$a_{30} = a + (30 – 1)d = a + 29d$$
$$a_{30} = 20 + 29(2) = 20 + 58 = 78\text{ seats}$$
Step 3: (ii) Total Seating Capacity ($S_{30}$)
Using the first and last term formula:
$$S_n = \frac{n}{2}(a + l)$$
where $a = 20$ and $l = a_{30} = 78$:
$$S_{30} = \frac{30}{2}(20 + 78) = 15(98) = 1470\text{ seats}$$
Alternatively, using the standard expansion formula:
$$S_{30} = \frac{30}{2}[2(20) + (30 – 1)(2)] = 15[40 + 58] = 15(98) = 1470$$
Final Answer:
(i) The 30th row contains $78\text{ seats}$.
(ii) The total seating capacity of the auditorium is $1470\text{ seats}$.
Question 11. [Personal Financial Savings Scheme]
A student begins a disciplined savings plan. She saves ₹$100$ in the first month and increases her monthly savings by ₹$50$ each subsequent month.
(i) How much does she save in the $12\text{th month}$?
(ii) In which month will her monthly savings reach ₹$1000$?
(iii) What is her total accumulated savings at the end of $1\text{ year}$ ($12\text{ months}$)?
Answer:
Step 1: Identify Given Data
Monthly savings follow an AP: $100, 150, 200, \dots$
- First term, $a = 100$
- Common difference, $d = 50$
Step 2: (i) Savings in the 12th Month ($a_{12}$)
$$a_{12} = a + (12 – 1)d = 100 + 11(50) = 100 + 550 = ₹650$$
Step 3: (ii) Month When Savings Reach ₹1000
Set $a_n = 1000$:
$$1000 = 100 + (n – 1)(50)$$
$$1000 – 100 = 50(n – 1)$$
$$900 = 50(n – 1)$$
$$n – 1 = \frac{900}{50} = 18 \implies n = 19$$
Her monthly savings will reach ₹$1000$ in the 19th month.
Step 4: (iii) Total Accumulated Savings in 1 Year ($S_{12}$)
$$S_n = \frac{n}{2}[2a + (n – 1)d]$$
$$S_{12} = \frac{12}{2}[2(100) + (12 – 1)(50)] = 6[200 + 550] = 6(750) = ₹4500$$
Final Answer:
(i) Savings in 12th month = ₹$650$.
(ii) Reaches ₹$1000$ in the $19\text{th month}$.
(iii) Total savings in 1 year = ₹$4500$.
[👉 Also Read: Class 10 Math Chapter 5 Arithmetic Progressions NCERT Solutions]
Master High-Yield Board FAQs (Rank Math Schema Ready)
What is the primary difference between a sequence and a series?
A sequence is an ordered list of numbers separated by commas where each term follows a specific generative rule (e.g., $2, 4, 6, 8$). A series is the expression obtained by adding the terms of a sequence together with addition signs (e.g., $2 + 4 + 6 + 8$).
How do you identify whether a sequence is an Arithmetic Progression?
To check if a sequence is an Arithmetic Progression (AP), subtract each term from the term that follows it ($a_{k+1} – a_k$). If this difference is constant across all consecutive pairs, the sequence is an AP, and the constant difference is the common difference $d$.
Can the common difference d of an AP be negative or zero?
Yes. The common difference $d$ can be positive, negative, or zero. If $d > 0$, the AP is strictly increasing; if $d < 0$, it is strictly decreasing; and if $d = 0$, all terms are identical, producing a constant progression.
What is the formula for the n-th term of an AP and what does each variable represent?
The formula for the $n$-th term is $a_n = a + (n – 1)d$. Here, $a_n$ represents the value of the term at position $n$, $a$ is the first term, $n$ is the positive integer position index ($n \in {1, 2, 3, \dots}$), and $d$ is the common difference.
Why is the factor in the AP formula (n – 1) instead of n?
The first term $a$ already starts at position $1$ without adding the common difference ($a_1 = a + 0 \cdot d$). To reach the $n$-th term from the first term, you add the common difference one fewer time than the total number of terms, which is $(n – 1)$ times.
What are triangular numbers and what is their general formula?
Triangular numbers are numbers that can be arranged in an equilateral triangular grid of dots. The $n$-th triangular number equals the sum of the first $n$ natural numbers, given by the formula $T_n = \frac{n(n + 1)}{2}$.
What is the recursive definition of the Fibonacci sequence?
The Fibonacci sequence is defined by the starting terms $F_1 = 1, F_2 = 1$ and the recursive relation $F_n = F_{n-1} + F_{n-2}$ for all $n \ge 3$. Each new term is the sum of the two terms directly preceding it ($1, 1, 2, 3, 5, 8, 13, \dots$).
What is the difference between an Arithmetic Progression and a Geometric Progression?
An Arithmetic Progression adds a fixed constant difference $d$ to get from one term to the next ($a_n = a + (n-1)d$). A Geometric Progression multiplies each term by a fixed non-zero ratio $r$ to obtain the subsequent term ($a_n = a \cdot r^{n-1}$).
Can the position index n in an AP ever be a fraction or negative?
No. The index $n$ represents the counting position of a term within an ordered list, so it must always be a positive integer ($n \in {1, 2, 3, \dots}$). If solving for $n$ gives a fraction or negative number, the value being tested is not part of that progression.
How do you find the sum of an AP when the first and last terms are known?
When the first term $a$ and last term $l$ are known, the sum of $n$ terms can be calculated using $S_n = \frac{n}{2}(a + l)$. This avoids having to compute the common difference $d$ first.
What happens when you sum two consecutive triangular numbers?
The sum of two consecutive triangular numbers always produces a perfect square: $T_{n-1} + T_n = \frac{(n-1)n}{2} + \frac{n(n+1)}{2} = \frac{n(n – 1 + n + 1)}{2} = \frac{2n^2}{2} = n^2$. For example, $T_2 + T_3 = 3 + 6 = 9 = 3^2$.
How do you find three numbers that form an Arithmetic Progression?
When three numbers in an AP need to be found and their sum is given, define them symmetrically as $(a – d)$, $a$, and $(a + d)$. Adding them cancels $d$ immediately ($3a = \text{Sum}$), simplifying the algebra.
What is Gauss’s method for summing the first n natural numbers?
Carl Friedrich Gauss noticed that writing a sum forward ($1 + 2 + \dots + n$) and backward ($n + (n-1) + \dots + 1$) yields $n$ pairs that each add up to $(n + 1)$. The double sum is $n(n + 1)$, so dividing by 2 gives the total: $S_n = \frac{n(n + 1)}{2}$.
What units should be written in progression real-life problems?
Units must match the physical quantities given in the problem statement (such as seats, bacteria, ₹, metres, or months). Omitting units on numerical answers in word problems typically leads to a half-mark deduction in CBSE examinations.
What are the key presentation steps to score 100% in AP questions?
To secure full marks: (1) state the given sequence and define the first term $a$ and common difference $d$, (2) state the general term formula ($a_n$) or sum formula ($S_n$) before substituting numbers, (3) write out intermediate algebraic steps clearly, and (4) state the final result with units in a concluding sentence.