NCERT Solutions Class 9 Mathematics Ganita Manjari Chapter 08 Predicting What Comes Next: Exploring Sequences and Progressions

Exercise 8.1

Question 1. Find the first five terms of the sequence in which the nth term is given by (i) tₙ = 3n − 4, (ii) tₙ = 2 − 5n, and (iii) tₙ = n² − 2n + 3 for n ≥ 1.
Answer: (i) When n = 1, 2, 3, 4, 5, the values become -1, 2, 5, 8, 11. The first five terms are: -1, 2, 5, 8, 11
(ii) When n = 1, 2, 3, 4, 5, the values become -3, -8, -13, -18, -23. The first five terms are: -3, -8, -13, -18, -23
(iii) When n = 1, 2, 3, 4, 5, the values become 2, 3, 6, 11, 18. The first five terms are: 2, 3, 6, 11, 18
In simple words: Use the formula to find each term by putting in the position number. For (i), multiply the position by 3 then subtract 4. For (ii), subtract 5 times the position from 2. For (iii), square the position, subtract twice the position, then add 3.

Exam Tip: Always substitute n = 1, 2, 3... carefully in order. Check your arithmetic — a small calculation error early on ruins the whole list.

Question 2. Find the 10th and 15th terms of the sequence tₙ = 5n − 3 for n ≥ 1.
Answer: 10th term: t₁₀ = 5(10) − 3 = 47
15th term: t₁₅ = 5(15) − 3 = 72
In simple words: Plug in the position number you want into the formula. Multiply by 5, then subtract 3.

Exam Tip: When finding a specific term, substitute the term number directly — do not calculate earlier terms first.

Question 3. Determine whether 97 and 172 are terms of the sequence tₙ = 5n − 3 for n ≥ 1.
Answer: For 97: Set tₙ = 97, giving 5n − 3 = 97. Solving gives n = 20, which is a natural number. So, 97 is the 20th term.
For 172: Set tₙ = 172, giving 5n − 3 = 172. Solving gives n = 35, which is a natural number. So, 172 is the 35th term.
In simple words: Put the value you are checking into the formula in place of tₙ. Solve for n. If n is a whole number, it is a term of the sequence.

Exam Tip: If n comes out as a fraction or negative, the number is not a term. Always state which term it is (e.g. "20th term") for full marks.

Question 4. Which term of the sequence tₙ = 5n − 3 for n ≥ 1 is 607?
Answer: Set tₙ = 607, giving 5n − 3 = 607. This leads to 5n = 610, so n = 122. Therefore, 607 is the 122nd term.
In simple words: Plug the number 607 into the formula and solve for n to find which position it is in the sequence.

Exam Tip: Always rearrange carefully and show your working step by step — examiners mark process as well as the final answer.

Question 5. A sequence is given by the recursive rule t₁ = −5, tₙ₊₁ = tₙ + 3 for n ≥ 1. Find the first five terms of the sequence. Is 52 a term of this sequence? If so, which term is it?
Answer: Starting with t₁ = −5, we add 3 to each term to find the next: t₂ = −2, t₃ = 1, t₄ = 4, t₅ = 7. The first five terms are: −5, −2, 1, 4, 7. This is an arithmetic sequence with first term a = −5 and common difference d = 3. Using the formula tₙ = −5 + (n − 1)3 = 3n − 8, setting tₙ = 52 gives 3n − 8 = 52, so n = 20. Therefore, 52 is the 20th term.
In simple words: Use the rule to build up the sequence term by term. Once you see the pattern is arithmetic, switch to the general formula to check if 52 belongs.

Exam Tip: Identify the type of sequence (arithmetic or geometric) early — this lets you write the closed-form formula and work faster.

Question 6. Let T₁ = 1, T₂ = 2, T₃ = 4, and Tₙ = Tₙ₋₁ + Tₙ₋₂ + Tₙ₋₃ for n ≥ 4. Find T₄, T₅, T₆, T₇, and T₈.
Answer: T₄ = 4 + 2 + 1 = 7
T₅ = 7 + 4 + 2 = 13
T₆ = 13 + 7 + 4 = 24
T₇ = 24 + 13 + 7 = 44
T₈ = 44 + 24 + 13 = 81
In simple words: Each new term is the sum of the three terms right before it. Keep adding the most recent three terms to build each new one.

Exam Tip: Write out each step clearly. Recursive sequences are error-prone if you use the wrong three (or two, or one) earlier terms — double-check which terms the rule calls for.

Exercise 8.2

Question 1. Find the 10th and 26th terms of the AP: 3, 8, 13, 18, ….
Answer: The first term is a = 3 and the common difference is d = 5. Using tₙ = a + (n − 1)d:
10th term: t₁₀ = 3 + 9 × 5 = 48
26th term: t₂₆ = 3 + 25 × 5 = 128
In simple words: Each term is 5 more than the one before. Multiply how many steps you take (9 for the 10th term, 25 for the 26th) by 5, then add to the first term.

Exam Tip: The formula uses (n − 1), not n. For the 10th term, you have taken 9 steps, so the exponent is 9, not 10.

Question 2. Which term of the AP: 21, 18, 15, … is −81? Also, is 0 a term of this AP? Give reasons for your answer.
Answer: The first term is a = 21 and the common difference is d = −3. The general term is tₙ = 21 + (n − 1)(−3) = 24 − 3n. For −81: Setting 24 − 3n = −81 gives n = 35, so −81 is the 35th term. For 0: Setting 24 − 3n = 0 gives n = 8, so 0 is the 8th term.
In simple words: Set the value you are checking equal to the formula, then solve for n. If n is a positive whole number, the value is a term.

Exam Tip: Always explain why a number is or is not a term by showing that n is (or is not) a natural number.

Question 3. Find the nth term of the AP: 11, 8, 5, 2, …. Write the recursive rule for this AP.
Answer: The first term is a = 11 and the common difference is d = −3. Using tₙ = a + (n − 1)d, the general term is tₙ = 14 − 3n. The recursive rule is: t₁ = 11, and tₙ = tₙ₋₁ − 3 for n ≥ 2.
In simple words: The explicit formula lets you jump to any term directly. The recursive rule tells you to subtract 3 from the previous term each time.

Exam Tip: Always provide both forms when asked — they show understanding of two different ways to describe sequences.

Question 4. An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term.
Answer: Let a be the first term and d the common difference. From the 3rd term: a + 2d = 12. From the 50th term: a + 49d = 106. Subtracting the first equation from the second: 47d = 94, so d = 2. Substituting back: a = 8. The 29th term is t₂₉ = 8 + 28 × 2 = 64.
In simple words: You have two pieces of information about the arithmetic sequence. Use both to set up two equations, then solve for a and d.

Exam Tip: Subtract equations strategically to eliminate one variable quickly. Always verify your answer by checking one of the original conditions.

Question 5. How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?
Answer: The 2-digit multiples of 3 form an arithmetic sequence: 12, 15, 18, …, 99. Here a = 12, d = 3, and the last term is 99. Using 99 = 12 + (n − 1)3, we get n = 30. So there are 30 such numbers. The sum is Sₙ = (30/2) × (12 + 99) = 15 × 111 = 1665.
In simple words: List all 2-digit numbers that are divisible by 3 in order - they form an arithmetic sequence. Find how many there are, then add them using the sum formula.

Exam Tip: The sum formula Sₙ = (n/2) × (first + last) is fast and less error-prone than adding all 30 numbers by hand.

Question 6. Harish started work at an annual salary of Rs 5,00,000 and received an increment of Rs 20,000 each year. After how many years did his income reach Rs 7,00,000?
Answer: His salary forms an arithmetic sequence starting at Rs 5,00,000 with common difference Rs 20,000. Setting up the equation Rs 7,00,000 = Rs 5,00,000 + (n − 1) × Rs 20,000, we get n − 1 = 10, so n = 11. Therefore, his income reached Rs 7,00,000 in the 11th year, which means after 10 years of work.
In simple words: Each year his pay goes up by Rs 20,000. Count how many years it takes his starting salary to increase by Rs 2,00,000.

Exam Tip: Be careful: the 11th year is reached after 10 years of work have passed. State the answer in the way the question asks for.

Question 7. A child arranges marbles in rows so that the first row has 1 marble, the second has 2 marbles, the third has 3, and so on up to 25 rows. How many marbles does the child use in all?
Answer: The total is 1 + 2 + 3 + ... + 25. Using the formula Sₙ = n(n + 1)/2 with n = 25: S₂₅ = 25 × 26 / 2 = 325. Therefore, the child uses 325 marbles in all.
In simple words: The marbles form a triangular pattern. Use the ready-made sum formula rather than adding 25 numbers individually.

Exam Tip: Memorise the sum formula Sₙ = n(n + 1)/2 — it saves time and applies to any sum of the first n natural numbers.

Exercise 8.3

Question 1. Find the 12th term of a GP with common ratio 2, whose 8th term is 192.
Answer: Given r = 2 and t₈ = ar⁷ = 192, we have a × 128 = 192, so a = 3/2. The 12th term is t₁₂ = ar¹¹ = (3/2) × 2¹¹ = 3 × 2¹⁰ = 3072.
In simple words: Find the first term using the given information. Then jump to the 12th term using the formula tₙ = arⁿ⁻¹.

Exam Tip: Write out the exponents carefully — for the 8th term it is r⁷, for the 12th it is r¹¹. Mismatched exponents are a common error.

Question 2. Find the 10th and nth terms of the GP: 5, 25, 125, ….
Answer: The first term is a = 5 and the common ratio is r = 5. The general term is tₙ = 5 × 5ⁿ⁻¹ = 5ⁿ. Therefore, the nth term is 5ⁿ, and the 10th term is t₁₀ = 5¹⁰ = 9765625.
In simple words: Each term is 5 times the previous one. The general formula simplifies neatly to just 5 to the power n.

Exam Tip: Look for simplifications in the formula — if the first term and ratio share a common base, the general term often simplifies to a single power.

Question 3. A sequence is given by the recursive rule t₁ = 2, tₙ₊₁ = 3tₙ − 2 for n ≥ 1. Which term of the sequence is 730?
Answer: Starting with t₁ = 2, we compute: t₂ = 3(2) − 2 = 4, t₃ = 3(4) − 2 = 10, t₄ = 3(10) − 2 = 28, t₅ = 3(28) − 2 = 82, t₆ = 3(82) − 2 = 244, t₇ = 3(244) − 2 = 730. Therefore, 730 is the 7th term.
In simple words: Apply the recursive rule repeatedly, multiplying by 3 and subtracting 2 each time, until you reach 730.

Exam Tip: For complex recursive rules, calculate step by step and keep a clear table — this avoids arithmetic errors and shows your work.

Question 4. Which term of the GP: 2, 6, 18, … is 4374? Write the explicit formula as well as the recursive formula for the nth term.
Answer: The first term is a = 2 and the common ratio is r = 3. The explicit formula is tₙ = 2 × 3ⁿ⁻¹. Setting 2 × 3ⁿ⁻¹ = 4374 gives 3ⁿ⁻¹ = 2187 = 3⁷, so n − 1 = 7 and n = 8. Therefore, 4374 is the 8th term. The recursive formula is: t₁ = 2, tₙ = 3tₙ₋₁ for n ≥ 2.
In simple words: Use the explicit formula to find which term equals 4374. The recursive formula shows you multiply by 3 each time to go to the next term.

Exam Tip: Always give both formulas when asked — one shows the "jump directly" approach, the other shows the "step by step" approach.

Question 5. A ball is dropped from a height of 80 metres. After hitting the ground, it bounces back to 60% of the height from which it fell. It continues bouncing in this way—each time rising to 60% of the previous height.
Answer (i): What height does the ball reach after the 5th bounce? After the 1st bounce the ball reaches 80 × 0.6 = 48 m. After the 2nd bounce: 48 × 0.6 = 28.8 m. After the 3rd bounce: 28.8 × 0.6 = 17.28 m. After the 4th bounce: 17.28 × 0.6 = 10.368 m. After the 5th bounce: 10.368 × 0.6 = 6.2208 m.
In simple words: Multiply the previous height by 0.6 five times in a row. Each bounce reaches only 60% of the height before it.

Exam Tip: Set up a geometric sequence with a = 80 and r = 0.6, then find the term you need. Multiplying step by step can introduce rounding errors — consider using the formula tₙ = 80 × (0.6)ⁿ directly.

Answer (ii): What is the total vertical distance the ball has travelled by the time it hits the ground for the 6th time? The ball first falls 80 m. After each of the first 5 bounces, it rises and then falls the same distance: 48 + 48 + 28.8 + 28.8 + 17.28 + 17.28 + 10.368 + 10.368 + 6.2208 + 6.2208. This simplifies to 80 + 2(48 + 28.8 + 17.28 + 10.368 + 6.2208) = 80 + 2(110.6688) = 80 + 221.3376 = 301.3376 m.
In simple words: The ball falls once at the start, then rises and falls after each bounce. Add the initial drop, then twice the sum of all bounce heights.

Exam Tip: Visualise the motion: down, up, down, up, down... The factor of 2 accounts for both the rise and fall at each bounce.

Question 6. Which term of the sequence 2, 2√2, 4, … is 128?
Answer: This is a GP with first term a = 2 and common ratio r = √2. Using tₙ = 2(√2)ⁿ⁻¹ and setting it equal to 128: (√2)ⁿ⁻¹ = 64. Since √2 = 2^(1/2), we have 2^((n-1)/2) = 2⁶. Comparing exponents: (n − 1)/2 = 6, so n − 1 = 12 and n = 13. Therefore, 128 is the 13th term.
In simple words: Convert √2 and 64 to powers of 2. Match the exponents to solve for n.

Exam Tip: When working with roots and powers, express everything as powers of the same base (here, base 2) — this makes exponent equations much simpler.

Question 7. Fig. 8.12 shows Stages 0 to 3 of the Sierpiński square carpet. Stage 0 of this fractal is a square sheet of paper. To construct Stage 1, each side of the square is trisected and the points of trisection of opposite sides are joined to obtain nine smaller squares. The centre square is then removed and the 8 smaller squares are retained, leaving a square hole in the centre. The same process is repeated on the eight smaller shaded squares to obtain Stage 2 and so on.
Answer (i): How many red squares are there in Stages 0 to 3? Stage 0 has 1 red square. Stage 1 has 8 red squares. Stage 2 has 8² = 64 red squares. Stage 3 has 8³ = 512 red squares.
In simple words: Each stage multiplies the number of squares by 8 because each square is divided into 9, and the centre one is removed, leaving 8.

Answer (ii): Can you predict the number of red squares in Stages 4 and 5? Stage 4 has 8⁴ = 4096 red squares. Stage 5 has 8⁵ = 32768 red squares.
In simple words: The sequence of red squares is a geometric progression with ratio 8. Multiply by 8 to get the next stage.

Answer (iii): Can you find a rule for the number of red squares at the nth stage? Write the explicit formula as well as the recursive formula for the number of red squares at any stage. The explicit formula is tₙ = 8ⁿ (where Stage 0 corresponds to n = 0). The recursive formula is: t₀ = 1, tₙ = 8tₙ₋₁ for n ≥ 1.
In simple words: The explicit formula lets you jump straight to any stage. The recursive formula shows that you multiply the previous stage by 8 each time.

Answer (iv): Suppose the area of the square in Stage 0 is 1 square unit. What is the area of the red region in Stages 1, 2 and 3? What will be the area of the red region in Stages 4 and 5? Find the explicit as well as the recursive formula for the area of the red region at the nth stage. What happens to this area as n, the number of stages, goes on increasing? At each stage, the square is divided into 9 parts and the centre is removed, so the remaining red area is 8/9 of the previous area. Stage 0: 1 square unit. Stage 1: 8/9. Stage 2: (8/9)² = 64/81. Stage 3: (8/9)³ = 512/729. Stage 4: (8/9)⁴ = 4096/6561. Stage 5: (8/9)⁵ = 32768/59049. The explicit formula is Aₙ = (8/9)ⁿ (where Stage 0 corresponds to n = 0). The recursive formula is: A₀ = 1, Aₙ = (8/9)Aₙ₋₁ for n ≥ 1. As n increases, the area approaches 0 but never reaches it exactly.
In simple words: The area shrinks by multiplying by 8/9 at each stage. Since 8/9 is less than 1, the areas get smaller and smaller, approaching zero as a limit.

Exam Tip: The Sierpiński carpet shows two geometric progressions at work: one (r = 8) grows the number of squares, the other (r = 8/9) shrinks the total area. Understanding this duality is key to the fractal concept.

End-of-Chapter Exercises

Question 1. Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
Answer: From the 11th term: a + 10d = 38. From the 16th term: a + 15d = 73. Subtracting gives 5d = 35, so d = 7. Substituting back: a = -32. The 31st term is t₃₁ = -32 + 30(7) = 178.
In simple words: Use the two given terms to set up equations for a and d. Once you know both, use the formula to jump to any term you need.

Exam Tip: Always solve the system of linear equations first — the examiners mark your ability to set up and solve equations, not just your final arithmetic.

Question 2. Determine the AP whose third term is 16 and whose 7th term exceeds the 5th term by 12.
Answer: From the third term: a + 2d = 16. The 7th term exceeds the 5th by 12, so (a + 6d) − (a + 4d) = 12, giving 2d = 12, thus d = 6. Substituting: a + 12 = 16, so a = 4. The AP is 4, 10, 16, 22, 28, ....
In simple words: Translate "exceeds by 12" as an equation. The difference between any two terms is 2d, so you can find d directly without needing the actual terms.

Exam Tip: The phrase "exceeds by" means subtraction: t₇ - t₅ = 12. This leads quickly to d = 6 without calculating the individual terms first.

Question 3. How many three-digit numbers are divisible by 7?
Answer: The smallest three-digit multiple of 7 is 105, and the largest is 994. The sequence 105, 112, 119, ..., 994 is an AP with a = 105 and d = 7. Setting 994 = 105 + (n − 1)7 gives 889 = 7(n − 1), so n − 1 = 127 and n = 128. Therefore, 128 three-digit numbers are divisible by 7.
In simple words: Identify the smallest and largest multiples in the range. Treat them as the first and last terms of an AP, then find how many terms there are.

Exam Tip: Always verify your smallest and largest terms before using them — a mistake here ruins the count.

Question 4. How many multiples of 4 lie between 10 and 250?
Answer: The smallest multiple of 4 greater than 10 is 12, and the largest less than 250 is 248. The sequence 12, 16, 20, ..., 248 has a = 12 and d = 4. Setting 248 = 12 + (n − 1)4 gives 236 = 4(n − 1), so n − 1 = 59 and n = 60. Therefore, 60 multiples of 4 lie between 10 and 250.
In simple words: Find the first and last multiples strictly within the given range, then count the terms in the arithmetic sequence they form.

Exam Tip: "Between" usually means the endpoints are included only if they satisfy the condition. Always check whether 10 or 250 themselves are multiples.

Question 5. Find a GP for which the sum of the first two terms is −4 and the fifth term is 4 times the third term.
Answer: Let the first term be a and the common ratio be r. From the first condition: a(1 + r) = −4. From the second condition: ar⁴ = 4ar², giving r² = 4, so r = 2 or r = −2. If r = 2: a(3) = −4, so a = −4/3. The GP is −4/3, −8/3, −16/3, −32/3, .... If r = −2: a(−1) = −4, so a = 4. The GP is 4, −8, 16, −32, .... Therefore, two possible GPs exist.
In simple words: Set up one equation from each condition. The second condition simplifies nicely to give you r directly — then find a from the first equation.

Exam Tip: Always check both values of r (if r² = 4 gives r = ±2) — a complete answer includes all solutions.

Question 6. Find all possible ways of expressing 100 as the sum of consecutive natural numbers.
Answer: The sum of consecutive integers from x to x + n - 1 is nx + n(n−1)/2. Setting this equal to 100 and solving for valid integer x, we find: 100 = 100 (1 term); 100 = 18 + 19 + 20 + 21 + 22 (5 terms); 100 = 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 (8 terms).
In simple words: Use the sum formula for an AP to find starting value x. For each choice of how many consecutive numbers (n), solve for x and check if it is a positive integer.

Exam Tip: Systematically try different values of n (1, 2, 3...). Not all will give whole-number starting values — only certain lengths work.

Question 7. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of the 2nd hour, 4th hour and nth hour?
Answer: The initial count is 30 and it doubles every hour, forming a GP with a = 30 and r = 2. At the end of the 2nd hour: 30 × 2² = 120 bacteria. At the end of the 4th hour: 30 × 2⁴ = 480 bacteria. At the end of the nth hour: 30 × 2ⁿ bacteria.
In simple words: Each hour, the population multiplies by 2. Use the formula tₙ = 30 × 2ⁿ, where n is the hour number.

Exam Tip: Be careful with indexing — at the end of the 1st hour, the count is 30 × 2¹. At the end of the nth hour, it is 30 × 2ⁿ (not 30 × 2^(n-1)).

Question 8. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Answer: From the first condition: (a + 3d) + (a + 7d) = 24, giving 2a + 10d = 24, so a + 5d = 12. From the second condition: (a + 5d) + (a + 9d) = 44, giving 2a + 14d = 44, so a + 7d = 22. Subtracting: 2d = 10, so d = 5. Substituting: a = -13. The first three terms are -13, -8, -3.
In simple words: Write the sum conditions as equations in a and d. Solve the system to find a and d, then write out the terms.

Exam Tip: Simplify immediately after forming equations — dividing by 2 early prevents arithmetic errors in later steps.

Question 9. Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000.
Answer: Using Sₙ = n(n + 1)/2, we need n(n + 1)/2 > 1000, or n(n + 1) > 2000. Testing: 44 × 45 = 1980, so S₄₄ = 990; 45 × 46 = 2070, so S₄₅ = 1035. Since 1035 > 1000, the smallest value is n = 45.
In simple words: Use trial and error around the approximate answer. Once you find two consecutive values of n that straddle 1000, pick the smaller n where the sum exceeds 1000.

Exam Tip: Estimate first: if n² ≈ 2000, then n ≈ 45. Start testing values around 45 to save time.

Question 10. Which term of the GP: 2, 8, 32, … is 131072? Write the explicit formula as well as the recursive formula for the nth term.
Answer: The first term is a = 2 and the common ratio is r = 4. The explicit formula is tₙ = 2 × 4ⁿ⁻¹. Setting 2 × 4ⁿ⁻¹ = 131072 gives 4ⁿ⁻¹ = 65536 = 4⁸, so n − 1 = 8 and n = 9. Therefore, 131072 is the 9th term. The recursive formula is: t₁ = 2, tₙ = 4tₙ₋₁ for n ≥ 2.
In simple words: Recognise that 65536 is a power of 4. Once you know 65536 = 4⁸, you can match exponents and solve for n.

Exam Tip: Familiarity with small powers of 2 and 4 (up to 2¹⁶ and 4⁸) helps you spot these without a calculator.

Question 11. The sum of the first three terms of a GP is 13/12 and their product is −1. Find the common ratio and the terms.
Answer: Let the three terms be a/r, a, ar. Their product is a³ = −1, so a = −1. The sum is −1/r − 1 − r = 13/12. Multiplying by −12r: 12 + 12r + 12r² = −13r, giving 12r² + 25r + 12 = 0. Factoring: (4r + 3)(3r + 4) = 0, so r = −3/4 or r = −4/3. If r = −3/4: terms are 4/3, −1, 3/4. If r = −4/3: terms are 3/4, −1, 4/3.
In simple words: Find a from the product condition. Substitute into the sum condition to get a quadratic in r. Solve for both values of r and list the corresponding terms.

Exam Tip: The quadratic 12r² + 25r + 12 = 0 factors nicely — look for factors of 12 × 12 = 144 that add to 25.

Question 12. If the 4th, 10th and 16th terms of a GP are x, y and z respectively, prove that x, y, z are in GP.
Answer: Let the first term be a and common ratio be r. Then: 4th term = ar³ = x, 10th term = ar⁹ = y, 16th term = ar¹⁵ = z. Computing the ratios: y/x = ar⁹ / ar³ = r⁶ and z/y = ar¹⁵ / ar⁹ = r⁶. Since y/x = z/y, the terms x, y, z form a GP.
In simple words: For three numbers to be in GP, the ratio of the second to the first must equal the ratio of the third to the second. Here, both ratios equal r⁶.

Exam Tip: A proof always ends with "Hence proved" or "Therefore proved" — do not omit this closing statement.

Question 13. The sum of the first three terms of a geometric progression is 26, and the sum of their squares is 364. Find the terms of the GP.
Answer: Let the terms be a, ar, ar². From the sum: a(1 + r + r²) = 26. From the sum of squares: a²(1 + r² + r⁴) = 364. Dividing the second equation by the square of the first and simplifying using the factorisation 1 + r² + r⁴ = (r² + r + 1)(r² − r + 1): (1 + r² + r⁴) / (1 + r + r²)² = 7/13. This leads to the quadratic 3r² − 10r + 3 = 0, giving r = 1/3 or r = 3. If r = 1/3: a = 18, so the terms are 18, 6, 2. If r = 3: a = 2, so the terms are 2, 6, 18.
In simple words: Set up two equations from the two conditions. Divide one by the other to eliminate a. Simplify and solve for r, then find a.

Exam Tip: Notice that both cases give the same three numbers in different orders — this is expected for symmetric problems.

Question 14. Suppose P₁ = 1, P₂ = 2 and for n > 2, Pₙ = P₁ + P₂ + … + Pₙ₋₁ + 1. Find the values of P₁, P₂, …, P₈. Can you find a simpler recursive formula for Pₙ? Can you give an explicit formula?
Answer: Computing step by step: P₃ = 1 + 2 + 1 = 4, P₄ = 1 + 2 + 4 + 1 = 8, P₅ = 1 + 2 + 4 + 8 + 1 = 16, P₆ = 32, P₇ = 64, P₈ = 128. The sequence 1, 2, 4, 8, 16, 32, 64, 128 is a geometric progression with ratio 2. The simpler recursive formula is: P₁ = 1, Pₙ = 2Pₙ₋₁ for n ≥ 2. The explicit formula is Pₙ = 2ⁿ⁻¹.
In simple words: Calculate the terms to spot the pattern. Once you see that each term is twice the previous, switch to the simpler recursive rule.

Exam Tip: Always look for hidden patterns — a complex-looking recursion can hide a simple geometric progression.

Question 15. Suppose W₁ = 1, W₂ = 2 and for n > 2, Wₙ = W₁ + W₂ + … + Wₙ₋₂ + 2. Find the values of W₁, W₂, …, W₈. Do you recognise this sequence?
Answer: Computing step by step: W₃ = 1 + 2 = 3, W₄ = 1 + 2 + 2 = 5, W₅ = 1 + 2 + 3 + 2 = 8, W₆ = 1 + 2 + 3 + 5 + 2 = 13, W₇ = 1 + 2 + 3 + 5 + 8 + 2 = 21, W₈ = 1 + 2 + 3 + 5 + 8 + 13 + 2 = 34. The sequence 1, 2, 3, 5, 8, 13, 21, 34 is the Virahanka-Fibonacci sequence, where each term is the sum of the two terms before it.
In simple words: Add all earlier terms except the last, then add 2. The result is a well-known sequence where each term equals the sum of the two before it.

Exam Tip: Recognising classic sequences (Fibonacci, triangular numbers, powers) shows deep understanding and saves time in exams.

Important Formulas of Class 9 Maths Ganita Manjari Chapter 8

These are the formulas every student must know before sitting any exam on Sequences and Progressions. Each one should be memorised with its meaning, not just its symbols.

What each variable means:

• a = first term of the AP or GP
• d = common difference (AP only)
• r = common ratio (GP only)
• n = position number of the term (always a natural number)
• tₙ = value of the term at position n
• Sₙ = sum of the first n natural numbers

Three things students always confuse:

1. In \( t_n = a + (n - 1)d \), the exponent on (n − 1) is 1, not n. Do not write \( t_n = a + nd \).
2. In \( t_n = ar^{n-1} \), the exponent is n − 1, not n. When n = 1, you must get \( t_1 = a \times r^0 = a \).
3. d is found by subtracting a term from the term after it (t₂ − t₁), not the term before it.

Summary of Chapter 8 Ganita Manjari - Sequences and Progressions

What is a Sequence?

A sequence is an ordered list of numbers where each number is called a term. Terms are labelled using subscripts - t₁ for the first, t₂ for the second and tₙ for the nth term. Sequences can be finite (fixed number of terms) or infinite (continuing indefinitely, shown by ...). The terms can be positive, negative, fractions or any real number. Some sequences follow a clear rule; others, like the prime numbers, do not have a simple pattern.

Explicit Rule vs Recursive Rule

An explicit rule expresses tₙ directly in terms of n - you can find any term without knowing others. For example, uₙ = 2n − 1 gives every odd number instantly. A recursive rule defines each term using the previous term. For example, t₁ = 1, tₙ = tₙ₋₁ + 3 for n ≥ 2. To use it, you must start from the first term and calculate step by step. The advantage is that it captures sequences which grow naturally from prior terms - like the Virahānka-Fibonacci sequence, where V₁ = 1, V₂ = 2, and Vₙ = Vₙ₋₁ + Vₙ₋₂ for n ≥ 3, giving 1, 2, 3, 5, 8, 13, 21, 34, ....

Arithmetic Progressions (AP)

An AP is a sequence where consecutive terms differ by a constant amount called the common difference (d). It can increase (d > 0), decrease (d < 0), or stay flat (d = 0). The nth term is \( t_n = a + (n - 1)d \). When the pairs (n, tₙ) are plotted on a graph, they always fall on a straight line - a visual signature of any AP.

Sum of First n Natural Numbers

Using Āryabhaṭa's method of writing the sum forwards and backwards, the formula \( S_n = \frac{n(n + 1)}{2} \) is derived. This same formula gives the nth triangular number, since each triangular number is the sum of natural numbers up to that position.

Geometric Progressions (GP)

A GP is a sequence where each term is obtained by multiplying the previous term by a fixed common ratio (r). The nth term is \( t_n = ar^{n-1} \). Unlike an AP, a GP grows exponentially - when plotted, its terms do not lie on a straight line but form a curve. If r > 1 the terms grow rapidly; if 0 < r < 1 they shrink toward zero; if r is negative the terms alternate in sign. The Sierpiński triangle is a striking real-world illustration: the number of black triangles at each stage follows a GP with r = 3 (growing fast), while the shaded area follows a GP with r = 3/4 (shrinking toward zero) - both operating simultaneously on the same figure.

Question 1. Find the 20th term of the AP: 3, 8, 13, 18, …
Answer: The first term is 3 and the common difference is 5. Using the formula tₙ = a + (n - 1)d, we get: t₂₀ = 3 + (20 - 1) × 5 = 3 + 95 = 98. So the 20th term is 98.
In simple words: Each term goes up by 5. To find the 20th term, add 5 nineteen times to the first term 3.

Exam Tip: Always identify a (first term) and d (common difference) first. These are the two key values needed for any AP formula.

Question 2. Find the nth term of the AP: 11, 8, 5, 2, …
Answer: Here the first term is 11 and the common difference is -3. Applying the formula tₙ = a + (n - 1)d: tₙ = 11 + (n - 1)(-3) = 11 - 3n + 3 = 14 - 3n. The general term is 14 - 3n.
In simple words: The terms go down by 3 each time. The formula 14 - 3n gives you any term in this list.

Exam Tip: Check your formula by substituting n = 1, 2, 3 - you should get the sequence numbers back.

Question 3. Which term of the AP: 21, 18, 15, … is -81?
Answer: The first term is 21 and the common difference is -3. Setting tₙ = -81 and using tₙ = a + (n - 1)d: 21 + (n - 1)(-3) = -81
(n - 1)(-3) = -102
n - 1 = 34
n = 35
The 35th term equals -81.
In simple words: We need to find which position in the sequence gives us -81. It takes 35 steps starting from 21 and going down by 3 each time to reach -81.

Exam Tip: Set tₙ equal to the given value and solve for n. The value of n tells you which term it is - it must be a positive whole number.

Question 4. An AP has 50 terms. The 3rd term is 12 and the last term is 106. Find the 29th term.
Answer: We use two conditions to find a and d. From the 3rd term: a + 2d = 12. From the 50th term: a + 49d = 106. Subtracting the first equation from the second: 47d = 94, so d = 2. Substituting back: a + 2(2) = 12, giving a = 8. Now we find the 29th term: t₂₉ = 8 + (29 - 1) × 2 = 8 + 56 = 64.
In simple words: Two facts about an AP let us find the starting term and the step size. Once we know these, we can find any term we need.

Exam Tip: When given two different terms, write two separate equations using tₙ = a + (n - 1)d. Subtract them to eliminate a and find d first.

Question 5. Harish started work with an annual salary of Rs. 5,00,000 and received an increment of Rs. 20,000 each year. After how many years did his income reach Rs. 7,00,000?
Answer: Harish's salary forms an AP with first term a = 5,00,000 and common difference d = 20,000. We need to find n such that tₙ = 7,00,000. Using tₙ = a + (n - 1)d: 5,00,000 + (n - 1) × 20,000 = 7,00,000
(n - 1) × 20,000 = 2,00,000
n - 1 = 10
n = 11
After 11 years (or at the start of the 11th year), his income reached Rs. 7,00,000.
In simple words: His salary goes up by the same amount every year. It takes 11 years for him to earn Rs. 7,00,000 a year.

Exam Tip: Always convert word problems into the AP formula carefully - identify what a, d, and tₙ represent in the context of the problem.

Question 6. How many 2-digit numbers are divisible by 3? Find their sum.
Answer: The 2-digit multiples of 3 form an AP: 12, 15, 18, …, 99. Here a = 12, d = 3, and the last term is 99. To find how many terms: 99 = 12 + (n - 1) × 3, which gives n - 1 = 29, so n = 30. There are 30 such numbers. The sum is calculated using Sₙ = (n/2)(first + last) = (30/2)(12 + 99) = 15 × 111 = 1,665.
In simple words: The 2-digit numbers divisible by 3 go from 12 to 99 with a gap of 3. There are 30 of them, and they add up to 1,665.

Exam Tip: Identify the first and last terms of the AP clearly. Use the last term to count how many terms exist, then apply the sum formula.

Question 7. Find the 10th term of the GP: 5, 25, 125, …
Answer: The first term is a = 5 and the common ratio is r = 5. Using the formula tₙ = arⁿ⁻¹: t₁₀ = 5 × 5⁹ = 5¹⁰ = 9,765,625. The 10th term is 9,765,625.
In simple words: Each term in this list is 5 times the previous one. The 10th term is found by multiplying 5 by itself 10 times total.

Exam Tip: For a GP, always find the common ratio first by dividing any term by the one before it. This ratio stays the same throughout.

Question 8. Find the nth term of the GP: 5, 25, 125, …
Answer: The first term is a = 5 and the common ratio is r = 5. Using tₙ = arⁿ⁻¹: tₙ = 5 × 5ⁿ⁻¹ = 5ⁿ. The general term is 5ⁿ.
In simple words: The nth term is simply 5 raised to the power n. This makes sense because 5¹ = 5, 5² = 25, 5³ = 125, and so on.

Exam Tip: Simplify powers when possible - combining the base a and r often leads to a cleaner formula.

Question 9. Is the sequence 2, 10, 50, 250, … a GP? Verify and find the common ratio.
Answer: To check if this is a GP, we verify that the ratio between consecutive terms is constant. The ratio 10/2 = 5, the ratio 50/10 = 5, and the ratio 250/50 = 5. Since all ratios are equal, this is a GP with common ratio r = 5. The nth term is tₙ = 2 × 5ⁿ⁻¹.
In simple words: Divide each term by the one before it. If you always get the same number, it is a GP. Here you always get 5.

Exam Tip: Check at least three ratios to be sure - never assume based on just one or two ratios.

Question 10. Is 308 a term of the sequence with explicit rule sₙ = 5n - 2?
Answer: We set sₙ = 308 and solve for n: 5n - 2 = 308, so 5n = 310, and n = 62. Since 62 is a natural number, 308 is the 62nd term of this sequence.
In simple words: Use the formula and replace sₙ with 308. If n comes out as a whole number, then 308 is in the sequence.

Exam Tip: If n is not a whole number, then that value is not part of the sequence - the formula only produces terms for positive integer values of n.

Question 11. Is 471 a term of the sequence sₙ = 5n - 2?
Answer: We set sₙ = 471 and solve for n: 5n - 2 = 471, so 5n = 473, and n = 94.6. Since 94.6 is not a natural number, 471 is not a term of this sequence.
In simple words: When we use the formula with 471, we get n = 94.6, which is not a whole number. So 471 does not appear in the sequence.

Exam Tip: Always check whether n is a whole number - if it has a decimal part, the value is definitely not in the sequence.

Question 12. Write both the explicit rule and the recursive rule for the AP: 4, 9, 14, 19, …
Answer: The first term is a = 4 and the common difference is d = 5. The explicit rule is: tₙ = 4 + (n - 1) × 5 = 5n - 1. The recursive rule is: t₁ = 4, and tₙ = tₙ₋₁ + 5 for n ≥ 2. The explicit rule lets you find any term directly, while the recursive rule defines each term based on the previous one.
In simple words: The explicit rule gives you any term directly using n. The recursive rule tells you how to get the next term from the current one.

Exam Tip: Both rules describe the same sequence - know when to use each one. Explicit rules are faster for large n; recursive rules show the step-by-step pattern.

Question 13. Find the sum: 25 + 26 + 27 + … + 58
Answer: This is a sequence of consecutive natural numbers. We use the formula for the sum of the first n natural numbers, Sₙ = n(n + 1)/2. The sum from 1 to 58 is S₅₈ = (58 × 59)/2 = 1,711, and the sum from 1 to 24 is S₂₄ = (24 × 25)/2 = 300. Therefore, the sum from 25 to 58 is: S₅₈ - S₂₄ = 1,711 - 300 = 1,411.
In simple words: To add numbers from 25 to 58, find the total from 1 to 58, then subtract the total from 1 to 24.

Exam Tip: When summing a range that doesn't start at 1, subtract the sum of terms before it from the sum of all terms up to the end.