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Sequences and Progressions Class 9 Worksheet Ch 8

Class IX Maths 🧩 Worksheet Ch 8: Predicting What Comes Next?: Exploring Sequences and Progressions NCERT book: Ganita Manjari
Updated 07 Sep 2026
"Predicting What Comes Next: Exploring Sequences and Progressions" — Class 09
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"Predicting What Comes Next: Exploring Sequences and Progressions"

Class 09 · Maths (Ganita Manjari) · Practice Worksheet with Solutions

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Section A · MCQ / Statement / Assertion–Reason / T-F (Q1–Q33)

1 Mark each
1
What is the nth term formula for a geometric progression where the first term is 'a' and the common ratio is 'r'?MCQ
  • a) tₙ = arⁿ⁻¹
  • b) tₙ = arⁿ
  • c) tₙ = a + nr
  • d) tₙ = a + (n − 1)r
✓ Correct Answer: (a) tₙ = arⁿ⁻¹
  1. This is the standard formula for the nₜ term of a geometric progression (GP): each term is the first term ‘a’ multiplied by the common ratio ‘r’, raised to the power (n−1).
2
Which term of the AP: 21, 42, 63, 84,... is 210?MCQ
  • a) 9th
  • b) 10th
  • c) 11th
  • d) 12th
✓ Correct Answer: (b) 10th
  1. Here a = 21, d = 21. Let Tₙ = 210: 21 + (n−1)(21) = 210.
  2. (n−1)(21) = 189 ⇒ n−1 = 9 ⇒ n = 10.
3
In an A.P., aₙ − aₙ₋₄ = 32. Its common difference is:MCQ
  • a) 4
  • b) 8
  • c) 4n
  • d) −8
✓ Correct Answer: (b) 8
  1. aₙ − aₙ₋₄ spans 4 common differences: aₙ − aₙ₋₄ = 4d.
  2. So 4d = 32 ⇒ d = 8.
4
Do the numbers 1², 5², 7², 73 ... form an A.P.? If yes, its next term will be:MCQ
  • a) Yes, 24
  • b) Yes, 11²
  • c) Yes, 97
  • d) No
✓ Correct Answer: (c) Yes, 97
  1. The list is 1, 25, 49, 73, ...
  2. a² − a₁ = 25−1 = 24, and a₃ − a₂ = 49−25 = 24 — the common difference is the same (24) throughout, so it is an A.P.
  3. Next term = 73 + 24 = 97.
5
The number of multiples of 4 that lie between 10 and 250 isMCQ
  • a) 55
  • b) 59
  • c) 62
  • d) 60
✓ Correct Answer: (d) 60
  1. The multiples of 4 between 10 and 250 are 12, 16, 20, ..., 248 — an A.P. with a = 12, d = 4.
  2. Let 248 be the nₜ term: 248 = 12 + (n−1)(4) ⇒ 4n = 240 ⇒ n = 60.
6
If k, 2k − 1 and 2k + 1 are three consecutive terms of an AP, the value of k isMCQ
  • a) −2
  • b) 3
  • c) −3
  • d) 6
✓ Correct Answer: (b) 3
  1. For consecutive AP terms, the middle term’s gap on both sides must match:
  2. (2k−1) − k = (2k+1) − (2k−1) ⇒ k − 1 = 2 ⇒ k = 3.
7
The 11th term of the A.P. −3, −12, 2, … isMCQ
  • a) 22
  • b) −48½
  • c) −38
  • d) 28
✓ Correct Answer: (a) 22
  1. Here a = −3, d = −12 − (−3) = 52.
  2. a₁₁ = a + 10d = −3 + 1052 = −3 + 25 = 22.
8
A specific bacterium population doubles every hour. If initially there are 100 bacteria, how many bacteria will there be after 5 hours?MCQ
  • a) 1600
  • b) 3200
  • c) 6400
  • d) 1000
✓ Correct Answer: (b) 3200
  1. This is a GP with first term a = 100 and common ratio r = 2.
  2. After 5 hours, population = t₆ (taking the initial count as t₁) = ar⁵ = 100 × 2⁵ = 100 × 32 = 3200.
9
If the sum of first p terms of an A.P. is ap² + bp, then its common difference is:MCQ
  • a) a
  • b) 3a + b
  • c) a + b
  • d) 2a
✓ Correct Answer: (d) 2a
  1. Tₚ = Sₚ − Sₚ₋₁ = (ap²+bp) − [a(p−1)²+b(p−1)] = 2ap − a + b.
  2. d = Tₚ − Tₚ₋₁ = (2ap−a+b) − [2a(p−1)−a+b] = 2a.
10
10th term of the A.P.: −12, −19, −26, … isMCQ
  • a) −75
  • b) −82
  • c) −65
  • d) 51
✓ Correct Answer: (a) −75
  1. Here a = −12, d = −19 − (−12) = −7.
  2. a₁₀ = a + 9d = −12 + 9(−7) = −12 − 63 = −75.
11
If the sum of the first 'n' terms of an AP is 2n² + 5n, then its common difference is:MCQ
  • a) 5
  • b) 4
  • c) 2
  • d) 7
✓ Correct Answer: (b) 4
  1. Sₙ = 2n²+5n. S₁ = a₁ = 2(1)+5(1) = 7. S₂ = 2(4)+5(2) = 18, so a₂ = S₂−S₁ = 11.
  2. Common difference d = a₂ − a₁ = 11 − 7 = 4.
12
If the 3rd term of a GP is 75 and the 5th term is 1875, what is the common ratio (r)?MCQ
  • a) 4
  • b) 6
  • c) 5
  • d) 3
✓ Correct Answer: (c) 5
  1. t₃ = ar² = 75 and t₅ = ar⁴ = 1875.
  2. Dividing: r² = 187575 = 25 ⇒ r = 5 (taking the positive root).
13
Which term of the A.P. 121, 117, 113, ... is its first negative term?MCQ
  • a) 33
  • b) 30
  • c) 31
  • d) 32
✓ Correct Answer: (d) 32
  1. Here a = 121, d = −4. We need the first n where aₙ < 0.
  2. 121 + (n−1)(−4) < 0 ⇒ 125 − 4n < 0 ⇒ n > 31.25.
  3. The smallest integer n satisfying this is n = 32.
14
The sum of first four terms of the G.P. 2, 6, 18,…, is:MCQ
  • a) 40
  • b) 37
  • c) 80
  • d) 58
✓ Correct Answer: (c) 80
  1. Common ratio r = 62 = 3. Fourth term = ar³ = 2(27) = 54.
  2. Sum of first four terms = 2 + 6 + 18 + 54 = 80.
15
The 4th term from the end of an AP −11, −8, −5, …., 49 isMCQ
  • a) 40
  • b) 37
  • c) 43
  • d) 58
✓ Correct Answer: (a) 40
  1. Here a = −11, d = 3, last term l = 49.
  2. 4th term from the end = l − (4−1)d = 49 − 3(3) = 49 − 9 = 40.
16
If x ≠ y and the sequences x, a₁, a₂, y and x, b₁, b₂, y each are in A.P., then a₂−a₁b₂−b₁ is ________.MCQ
  • a) 1
  • b) 34
  • c) 32
  • d) 23
✓ Correct Answer: (a) 1
  1. For x, a₁, a₂, y in A.P. (4 terms, 3 gaps): common difference d = y−x3, so a₂−a₁ = d = y−x3.
  2. Similarly for x, b₁, b₂, y: b₂−b₁ = y−x3 also.
  3. So a₂−a₁b₂−b₁ = [y−x3] / [y−x3] = 1.
17
If the sum of first n terms of an A.P. is given by Sₙ = n2(3n + 1), then the first term of the A.P. isMCQ
  • a) 52
  • b) 4
  • c) 2
  • d) 32
✓ Correct Answer: (c) 2
  1. First term = S₁ (sum of just the first term).
  2. S₁ = 12(3(1)+1) = 12(4) = 2.
18
In a G.P., common ratio = 2, first term = 3 and last term = 96.
Statement (1): The number of terms in this G.P. = 96 − 3.
Statement (2): a : arⁿ⁻¹ = 3 : 96Statement-Based
  • a) Both the statements are true.
  • b) Both the statements are false.
  • c) Statement 1 is true, and statement 2 is false.
  • d) Statement 1 is false, and statement 2 is true.
✓ Correct Answer: (d) Statement 1 is false, and statement 2 is true.
  1. Using Tₙ = arⁿ⁻¹: 3×2ⁿ⁻¹ = 96 ⇒ 2ⁿ⁻¹ = 32 = 2⁵ ⇒ n−1 = 5 ⇒ n = 6.
  2. Statement 1 claims n = 96−3 = 93, which does not match the correct n = 6 — so Statement 1 is false.
  3. Statement 2 correctly expresses the ratio of first term to last term as a : arⁿ⁻¹ = 3 : 96 — this is true by definition.
19
The sum of the GP 13, 19, 127, 181, … to n terms isMCQ
  • a) None of these
  • b) 23[13ⁿ − 1]
  • c) 12[13ⁿ − 1]
  • d) 12[1 − 13ⁿ]
✓ Correct Answer: (d) 12[1 − 13ⁿ]
  1. Here a = 13, r = 13 (r < 1), so Sₙ = a1−rⁿ1−r.
  2. Sₙ = 13[1−13ⁿ] / 23 = 12[1 − 13ⁿ].
20
The third term of GP is 4, the product of the first five terms isMCQ
  • a) 256
  • b) 1024
  • c) 512
  • d) 612
✓ Correct Answer: (b) 1024
  1. Let the first five terms be ar², ar, a, ar, ar², with third term = a = 4.
  2. Product of first five terms = ar²ar(a)(ar)(ar²) = a⁵ = 4⁵ = 1024.
21
If Sₙ denotes the sum of n terms of an A.P. with first term a and common difference d such that SₓSₖₓ is independent of x, thenMCQ
  • a) d = −a
  • b) a = 2d
  • c) d = 2a
  • d) d = a
✓ Correct Answer: (c) d = 2a
  1. Sₓ = x2[2a+(x−1)d] and Sₖₓ = kx2[2a+(kx−1)d].
  2. SₓSₖₓ = [(2a−d)+xd] / [k{(2a−d)+kxd}].
  3. This ratio is independent of x only when the constant term (2a−d) vanishes, i.e. 2a − d = 0 ⇒ d = 2a.
22
Which of the following is not an A.P.?MCQ
  • a) 2, 4, 8, 16, ...
  • b) −1.2, −3.2, −5.2, −7.2, ...
  • c) 2, 52, 3, 72, ...
  • d) a, 2a, 3a, 4a, ...
✓ Correct Answer: (a) 2, 4, 8, 16, ...
  1. Check consecutive differences in 2, 4, 8, 16: 4−2=2, 8−4=4, 16−8=8 — not constant.
  2. Since the common difference is not the same throughout, this sequence is not an A.P. (it is in fact a GP).
23
The number of terms of the A.P. 5, 8, 11, 14, ... to be taken so that the sum is 258 isMCQ
  • a) 10
  • b) 14
  • c) 12
  • d) 16
✓ Correct Answer: (c) 12
  1. Here a=5, d=3. Sₙ = n2[2a+(n−1)d] = 258.
  2. n2[10+3n−3] = 258 ⇒ n(3n+7) = 516 ⇒ 3n²+7n−516 = 0.
  3. Factoring: (n−12)(3n+43) = 0 ⇒ n = 12 (rejecting the negative or fractional root).
24
The sum of 20 terms of the G.P. 10, 20, 40,… is:MCQ
  • a) 10(2²²−1)
  • b) 10(2²¹−1)
  • c) 10(2¹⁹−1)
  • d) 10(2²⁰−1)
✓ Correct Answer: (d) 10(2²⁰−1)
  1. Here a = 10, r = 2010 = 2. Since |r| > 1, use Sₙ = arⁿ−1r−1.
  2. S₂₀ = 102²⁰−12−1 = 10(2²⁰−1).
25
If Sₙ denotes the sum of the first n terms of an A.P. Then, S₃ₙ : (S₂ₙ−Sₙ) isMCQ
  • a) Arithmetic
  • b) Geometric
  • c) 3
  • d) n
✓ Correct Answer: (c) 3
  1. Sₙ = n2[2a+(n−1)d], S₂ₙ = n[2a+(2n−1)d], S₃ₙ = 3n2[2a+(3n−1)d].
  2. S₂ₙ−Sₙ simplifies to n[3a+(3n−2)d]/1 (a middle-AP-block sum), and dividing S₃ₙ by this difference works out to the constant ratio 3 (a standard result for these three consecutive block sums of an AP).
26
Given a₁, a₂, a₃, … and b₁, b₂, b₃, … are real numbers such that a₁−b₁ = a₂−b₂ = a₃−b₃ = … are all equal. a₁−b₁, a₂−b₂, a₃−b₃… forms a ________ progression.MCQ
  • a) Arithmetic (d = 1)
  • b) Geometric (r = 1)
  • c) Geometric (r < 1)
  • d) Arithmetic (d = 0)
✓ Correct Answer: (d) Arithmetic (d = 0)
  1. Since all the terms aₙ−bₙ are equal (constant), consecutive terms have zero difference.
  2. A sequence of equal terms is a (trivial) Arithmetic Progression with common difference d = 0.
27
Statement I: The sum of the series 2 + 6 + 18 + 54 + … + 4374 is 6560.
Statement II: If a and r are the first term and common ratio of a GP, then sum of first n terms of this GP is given by Sₙ = a1−rⁿ1−r, when r < 1.Statement-Based
  • a) Statement I is false and Statement II is true.
  • b) Both the Statements are false.
  • c) Statement I is true and Statement II is false.
  • d) Both the Statements are true.
✓ Correct Answer: (d) Both the Statements are true.
  1. Statement I: This is a GP with a=2, r=3, last term l=4374. Sum = lr−ar−1 = 4374×3−23−1 = 131202 = 6560. — True.
  2. Statement II correctly states the standard GP sum formula for r < 1 — True.
  3. (Note Statement II is a general true fact, even though Statement I’s GP actually has r=3>1; the two statements are independently evaluated for truth.)
28
Given x, y, z are in A.P.
Assertion (A): z, y, x are in A.P.
Reason (R): The terms of an A.P. taken in reverse order also form an A.P.Assertion-Reason
  • a) Both A and R are true and R is the correct explanation of A.
  • b) Both A and R are true but R is not the correct explanation of A.
  • c) A is true but R is false.
  • d) A is false but R is true.
✓ Correct Answer: (a) Both A and R are true and R is the correct explanation of A.
  1. Since x, y, z are in A.P., y − x = z − y, i.e. x + z = 2y.
  2. For z, y, x to be in A.P., we need y−z = x−y, i.e. x+z = 2y — the same condition, so it holds. Hence z, y, x are indeed in A.P. (with the sign of d reversed) — Assertion is true.
  3. The Reason correctly generalises this: reversing the order of any A.P. still gives an A.P. (same common difference, negated) — true, and it directly explains the Assertion.
29
Tₙ = 4 − 2n is the nth term of an A.P.
Assertion (A): This A.P. will have all terms negative after the 2nd term.
Reason (R): The common difference is negative.Assertion-Reason
  • a) Both A and R are true and R is the correct explanation of A.
  • b) Both A and R are true but R is not the correct explanation of A.
  • c) A is true but R is false.
  • d) A is false but R is true.
✓ Correct Answer: (a) Both A and R are true and R is the correct explanation of A.
  1. T₁=2, T₂=0, T₃=−2, T₄=−4, … — indeed every term from the 3rd onward (i.e. after the 2nd term) is negative. Assertion is true.
  2. Common difference d = Tₙ − Tₙ₋₁ = −2, which is negative — Reason is true, and a negative d is exactly why the terms keep decreasing into negative values, correctly explaining the Assertion.
30
Assertion (A): Common difference of the A.P. 5, 1, −3, −7 … is 4.
Reason (R): Common difference of the A.P. a₁, a₂, a₃ … aₙ is obtained by d = aₙ − aₙ₋₁.Assertion-Reason
  • a) Both A and R are true and R is the correct explanation of A.
  • b) Both A and R are true but R is not the correct explanation of A.
  • c) A is true but R is false.
  • d) A is false but R is true.
✓ Correct Answer: (d) A is false but R is true.
  1. Actual common difference: d = 1 − 5 = −4, not 4 — so Assertion is false.
  2. The Reason correctly states the general formula d = aₙ − aₙ₋₁ — this is true.
31
State whether the given statement is True or False:
(a) Sum of first n positive integers is given by Sₙ = nn+12.
  1. This is the standard, well-known formula for the sum of the first n positive integers (1+2+3+…+n).
  2. It can be derived by pairing the first and last terms: (1+n), (2+n−1), etc., each summing to (n+1), with n2 such pairs.
  3. So the statement is correctly stated.
✓ (a) True
32
Fill in the blanks:
(a) The sum of the AP, 1 + 2 + 3 + 4 + 5 + 6 + …… 10 is ________.
  1. This is the sum of the first 10 positive integers: Sₙ = nn+12 with n = 10.
  2. S₁₀ = 10112 = 55.
✓ (a) 55
33
Fill in the blanks:
(a) If Sₙ denotes the sum of first n terms of an A.P., then S₂ − S₁ = ________.
  1. S₁ = a₁ = a (the first term itself). S₂ = a₁ + a₂.
  2. So S₂ − S₁ = a₂, the second term of the A.P., which equals a + d.
✓ (a) a₂ (i.e. a + d)

Section B · Short Answer — AP & GP (Q34–Q50)

2 Marks each
34
Find the 8th term from the end of the A.P. 7, 10, 13, ..., 184.
  1. Here last term l = 184, common difference d = 10 − 7 = 3.
  2. n-th term from the end = l − (n−1)d, so 8th term from end = 184 − (8−1)(3) = 184 − 21.
✓ 8th term from the end = 163
35
A man receives Rs. 60 for the first week and Rs. 3 more each week than the preceding week. How much does he earn by the 20th week?
  1. This forms an A.P.: 60, 63, 66, … up to 20 terms, with a = 60, d = 3.
  2. Total earning = S₂₀ = 202[2(60) + 19(3)] = 10[120 + 57] = 10(177).
✓ Total earning by the 20th week = Rs. 1770
36
Find the sum: 32 + 30 + 28 + ... + 10.
  1. Here a = 32, d = 30−32 = −2, last term l = 10.
  2. Find n: l = a+(n−1)d ⇒ 10 = 32+(n−1)(−2) ⇒ (n−1)(−2) = −22 ⇒ n−1 = 11 ⇒ n = 12.
  3. Sₙ = n2(a+l) = 122(32+10) = 6 × 42.
✓ Sum = 252
37
The sum of the 5th and the 7th terms of an AP is 52 and the 10th term is 46. Find the AP.
  1. a₅ + a₇ = 52 ⇒ (a+4d)+(a+6d) = 52 ⇒ 2a+10d = 52 ⇒ a+5d = 26 — (1)
  2. a₁₀ = 46 ⇒ a+9d = 46 — (2)
  3. From (1): a = 26−5d. Substitute into (2): 26−5d+9d = 46 ⇒ 4d = 20 ⇒ d = 5.
  4. Then a = 26 − 5(5) = 1.
✓ The AP is 1, 6, 11, 16, … (a = 1, d = 5)
38
Find the value of x such that x + 9, x − 6 and 4 are three consecutive terms of a G.P.
  1. For a GP, the ratio of consecutive terms is equal: x−6x+9 = 4x−6.
  2. Cross-multiplying: (x−6)² = 4(x+9) ⇒ x²−12x+36 = 4x+36.
  3. x² − 16x = 0 ⇒ x(x−16) = 0 ⇒ x = 0 or x = 16.
✓ x = 0 or x = 16
39
Is −150 a term of the AP 17, 12, 7, 2 .....?
  1. Here a = 17, d = 12−17 = −5. Suppose aₙ = −150.
  2. 17 + (n−1)(−5) = −150 ⇒ (n−1)(−5) = −167 ⇒ n−1 = 1675 = 33.4.
  3. Since n must be a whole number and 33.4 is not an integer, −150 is not a term of this A.P.
✓ No, −150 is not a term of the AP
40
Find the sum of first 8 multiples of 3.
  1. First 8 multiples of 3: 3, 6, 9, ..., 24 = 3(1+2+3+…+8).
  2. Sum = 3 × [8×92] = 3 × 36.
✓ Sum of first 8 multiples of 3 = 108
41
Find the sum of all three-digit natural numbers which are divisible by 13.
  1. Three-digit multiples of 13: 104, 117, 130, ..., 988 — an A.P. with a=104, d=13, l=988.
  2. Find n: 988 = 104+(n−1)(13) ⇒ (n−1)(13) = 884 ⇒ n−1 = 68 ⇒ n = 69.
  3. Sum = n2(a+l) = 692(104+988) = 692(1092) = 69 × 546.
✓ Sum = 37,674
42
Find the 12th term of a GP with common ratio 2, whose 8th term is 192.
  1. t₈ = ar₄ = 192, with r = 2: a(2₄) = 192 ⇒ 128a = 192 ⇒ a = 32.
  2. t₁₂ = ar¹¹ = 32(2¹¹) = 32(2048) = 3 × 1024.
✓ 12th term = 3072
43
The fourth term of a G.P. is the square of its second term and the first term is −3. Find its 7th term.
  1. Given a₄ = (a₂)², i.e. ar³ = (ar)² = a²r².
  2. With a = −3: (−3)r³ = (−3r)² = 9r² ⇒ −3r³ = 9r² ⇒ 9r²+3r³ = 0 ⇒ 3r²(3+r) = 0.
  3. Since r ≠ 0, r = −3.
  4. The 7th term t₅ = ar⁶ = (−3)(−3)⁶ = −3 × 729.
✓ 7th term = −2187
44
Find the number of all integers between 2 and 100, divisible by 3.
  1. Multiples of 3 between 2 and 100: 3, 6, 9, ..., 99 — an A.P. with a=3, d=3, last term=99.
  2. 99 = 3+(n−1)(3) ⇒ 96 = (n−1)(3) ⇒ n−1 = 32 ⇒ n = 33.
✓ There are 33 such integers
45
Find the sum of first 6 terms of the G.P. 0.1, 0.01, 0.001, ...
  1. Here a = 0.1, r = 0.010.1 = 0.1 (r < 1), n = 6.
  2. Sₙ = a1−rⁿ1−r = 0.1[1−(0.1)⁶]/(1−0.1) = 0.10.9999990.9.
  3. = 19(0.999999) = 0.111111.
✓ S₆ = 0.111111
46
Check whether 301 is a term of the given list of numbers: 5, 11, 17, 23,...?
  1. Check the differences: 11−5=6, 17−11=6, 23−17=6 — constant, so this is an A.P. with a=5, d=6.
  2. Suppose 301 is the n-th term: 301 = 5+(n−1)(6) ⇒ 301 = 6n−1 ⇒ 6n = 302 ⇒ n = 3026 = 1513.
  3. Since n is not a whole number, 301 is not a term of this AP.
✓ No, 301 is not a term of the given AP
47
If a, a² + 2 and a³ + 10 are in G.P., then find the value(s) of a.
  1. For a GP: a²+2a = a³+10a²+2.
  2. Cross-multiplying: (a²+2)² = a(a³+10) ⇒ a⁴+4a²+4 = a⁴+10a.
  3. 4a² − 10a + 4 = 0 ⇒ 2a² − 5a + 2 = 0 ⇒ 4a²−8a−2a+4=0 ⇒ 4a(a−2)−2(a−2)=0 ⇒ (4a−2)(a−2)=0.
  4. So a = 12 or a = 2.
✓ a = 12 or a = 2
48
Show that the progression 6, 5½, 5, 4½, 4, … is an A.P. Write down its:
i. first term
ii. common difference
iii. 9th term.
  1. The series is 6, 5.5, 5, 4.5, 4, … Check differences: 5.5−6=−0.5, 5−5.5=−0.5, 4.5−5=−0.5, 4−4.5=−0.5 — all equal.
  2. So the progression is indeed an A.P. with common difference −0.5.
  3. (i) First term a = 6. (ii) Common difference d = −0.5.
  4. (iii) a₉ = a + 8d = 6 + 8(−0.5) = 6 − 4 = 2.
✓ a = 6, d = −0.5, 9th term = 2
49
Find the 25th term of the A.P 5, 4½, 4, 3½, 3, …
  1. Here a = 5, d = 4.5 − 5 = −0.5 (i.e. −12).
  2. a₂₅ = a + 24d = 5 + 24−12 = 5 − 12.
✓ 25th term = −7
50
If the nth term of a progression is (4n − 10), show that it is an A.P. Find its 16th term.
  1. T₁ = 4(1)−10 = −6, T₂ = 4(2)−10 = −2, T₃ = 4(3)−10 = 2, T₄ = 4(4)−10 = 6.
  2. Differences: −2−(−6)=4, 2−(−2)=4, 6−2=4 — all equal, so it is indeed an A.P. with a=−6, d=4.
  3. T₁₆ = a + 15d = −6 + 15(4) = −6 + 60.
✓ It is an A.P.; 16th term = 54

Section C · Short Answer — AP & GP (Q51–Q65)

3 Marks each
51
Find an AP whose 4th term is 9 and the sum of its 6th and 13th terms is 40.
  1. a₄ = 9 ⇒ a + 3d = 9 — (1)
  2. a₆ + a₁₃ = 40 ⇒ (a+5d)+(a+12d) = 40 ⇒ 2a+17d = 40 — (2)
  3. Multiply (1) by 2: 2a + 6d = 18 — (3). Subtract (3) from (2): 11d = 22 ⇒ d = 2.
  4. From (1): a = 9 − 3(2) = 3.
✓ The AP is 3, 5, 7, 9, … (a = 3, d = 2)
52
If a, b and c in G.P., prove that: log aⁿ, log bⁿ and log cⁿ are in A.P.
  1. Since a, b, c are in G.P., b² = ac.
  2. Taking log on both sides: log(b²) = log(ac) ⇒ 2 log b = log a + log c ⇒ log b + log b = log a + log c.
  3. Rearranging: log b − log a = log c − log b, which is the condition for log a, log b, log c to be in A.P.
  4. Multiplying throughout by n (a constant): n·log a, n·log b, n·log c are also in A.P., i.e. log aⁿ, log bⁿ, log cⁿ are in A.P.
✓ log aⁿ, log bⁿ, log cⁿ are in A.P. — proved.
53
Find the sum of G.P.: 1 − 13 + 13² − 13³ + …. to n terms.
  1. Here a = 1, r = −13 (since |r| < 1, use Sₙ = a1−rⁿ1−r).
  2. Sₙ = [1−−13ⁿ] / [1−−13] = [1−−13ⁿ] / 43.
✓ Sₙ = 34[1 − −13ⁿ]
54
The first and the last terms of an A.P. are 34 and 700 respectively. If the common difference is 18, how many terms are there and what is their sum?
  1. a = 34, d = 18, l = 700. Find n: 700 = 34+(n−1)(18) ⇒ (n−1)(18) = 666 ⇒ n−1 = 37 ⇒ n = 38.
  2. Sum = n2(a+l) = 382(34+700) = 19 × 734.
✓ Number of terms = 38; Sum = 13,946
55
Which term of A.P. 5, 15, 25 …… will be 130 more than its 31st term?
  1. Here a = 5, d = 10. a₃₁ = a + 30d = 5 + 300 = 305.
  2. We need tₙ = 305 + 130 = 435. So 5 + (n−1)(10) = 435 ⇒ (n−1)(10) = 430 ⇒ n−1 = 43 ⇒ n = 44.
✓ The 44th term is 130 more than the 31st term
56
Find the sum of the first 40 positive integers divisible by 6.
  1. First 40 multiples of 6: 6, 12, 18, ... (a=6, d=6, n=40).
  2. Sₙ = n2[2a+(n−1)d] = 402[2(6)+39(6)] = 20[12+234] = 20(246).
✓ Sum = 4920
57
Find the nth term and write the recursive rule for the AP: 20, 17, 14, 11, …
  1. Here a = 20, d = 17−20 = −3.
  2. Explicit rule: tₙ = a+(n−1)d = 20+(n−1)(−3) = 20−3n+3 = 23−3n.
  3. Recursive rule: t₁ = 20, and tₙ = tₙ₋₁ − 3 for n ≥ 2.
✓ tₙ = 23 − 3n; recursive: t₁=20, tₙ=tₙ₋₁−3 (n≥2)
58
Find the G.P. whose first term is 64 and next term is 32.
  1. a = 64, second term = ar = 32 ⇒ r = 3264 = 12.
  2. G.P. = a, ar, ar², ar³, … = 64, 32, 6414, 6418, … = 64, 32, 16, 8, …
✓ The G.P. is 64, 32, 16, 8, … (r = 12)
59
Which term of the AP 15, 30, 45, 60, ... is 300? Hence, find the sum of all the terms of the AP.
  1. a = 15, d = 15. Let aₙ = 300: 15+(n−1)(15) = 300 ⇒ (n−1)(15) = 285 ⇒ n−1 = 19 ⇒ n = 20.
  2. So 300 is the 20th term. Sum of all 20 terms: S₂₀ = 202(15+300) = 10 × 315.
✓ 300 is the 20th term; Sum of all terms = 3150
60
Find the sum of first 20 terms of an A.P., in which 3rd term is 7 and 7th term is two more than thrice of its 3rd term.
  1. a₃ = 7 ⇒ a+2d = 7. a₇ = 3a₃+2 = 23, so a+6d = 23.
  2. Subtracting: 4d = 16 ⇒ d = 4. Then a = 7−2(4) = −1.
  3. S₂₀ = 202[2(−1)+19(4)] = 10[−2+76] = 10(74).
✓ Sum of first 20 terms = 740
61
If the sum of the first 14 terms of an A.P. is 1050 and its first term is 10, then find the 21st term of the A.P.
  1. S₁₄ = 142[2(10)+13d] = 1050 ⇒ 7(20+13d) = 1050 ⇒ 20+13d = 150 ⇒ 13d = 130 ⇒ d = 10.
  2. a₂₁ = a + 20d = 10 + 20(10) = 10 + 200.
✓ 21st term = 210
62
The eighth term of an A.P. is half of its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15th term.
  1. a₈ = ½a₂ ⇒ 2(a+7d) = a+d ⇒ a + 13d = 0 — (i)
  2. a₁₁ − ⅓a₄ = 1 ⇒ 3(a+10d) = a+3d+3 ⇒ 2a+27d = 3 — (ii)
  3. From (i)×2: 2a+26d = 0. Subtracting from (ii): d = 3. Then from (i): a = −39.
  4. a₁₅ = a+14d = −39+14(3) = −39+42.
✓ 15th term = 3
63
The 7th term of an AP is −4 and its 13th term is −16. Find the AP.
  1. a₇ = a+6d = −4 — (1); a₁₃ = a+12d = −16 — (2)
  2. Subtracting (1) from (2): 6d = −12 ⇒ d = −2.
  3. From (1): a − 12 = −4 ⇒ a = 8.
✓ The AP is 8, 6, 4, 2, 0, … (a=8, d=−2)
64
Find the values of t₁, t₅, and t₁₂ for the sequence whose explicit formula is tₙ = (−1)ⁿ · (2n + 1).
  1. For n=1: t₁ = (−1)¹(2(1)+1) = −1 × 3 = −3.
  2. For n=5: t₅ = (−1)⁵(2(5)+1) = −1 × 11 = −11.
  3. For n=12: t₁₂ = (−1)¹²(2(12)+1) = 1 × 25 = 25.
✓ t₁ = −3, t₅ = −11, t₁₂ = 25
65
If the mth term of an A.P. is 1n and nth term be 1m, then show that its (mn)th term is 1.
  1. a₞ = a+(m−1)d = 1n — (i); aₙ = a+(n−1)d = 1m — (ii)
  2. Subtracting (ii) from (i): (m−n)d = 1n − 1m = m−nmn ⇒ d = 1mn.
  3. Substituting d back into (i): a + m−1mn = 1n ⇒ a = 1n − m−1mn = m − m + 1mn = 1mn.
  4. So a₞ₙ = a + (mn−1)d = 1mn + mn−1mn = mnmn = 1.
✓ (mn)th term = 1 — proved.

Section D · Case Study Based (Q66–Q70)

4 Marks each
66
Case Study: In an equilateral triangle of side 10 cm, equilateral triangles of side 1 cm are formed as shown in the figure, such that there is one triangle in the first row, three triangles in the second row, five triangles in the third row, and so on (up to the 10th row). Answer the following questions using Arithmetic Progression.
Rows of small triangles inside the large triangle (Q66)
Rows of small triangles inside the large triangle (Q66)
  • How many triangles will be there in the bottommost row? (1)
  • How many triangles will be there in the fourth row from the bottom? (1)
  • a. Find the total number of triangles of side 1 cm each till the 8th row. (2)
    OR — b. How many more triangles are there from the 5th row to the 10th row than in the first 4 rows? Show working. (2)
  1. The number of triangles per row forms the A.P. 1, 3, 5, 7, … (a = 1, d = 2), row 1 at the top down to row 10 at the bottom.
  2. (i) Bottommost row = 10th row: t₁₀ = a+9d = 1+9(2) = 1+18 = 19 triangles.
  3. (ii) The 4th row from the bottom is row (10−4+1) = row 7 from the top: t₄ from the end = l − (4−1)d = 19 − 3(2) = 19 − 6 = 13 triangles.
  4. (iii-a) Total triangles up to the 8th row = S₈ = 82[2(1)+7(2)] = 4[2+14] = 4(16) = 64.
  5. OR (iii-b): Triangles in first 4 rows: S₄ = 42[2(1)+3(2)] = 2(8) = 16.
  6. Triangles from 5th to 10th row = S₁₀ − S₄ = 102[2(1)+9(2)] − 16 = 5(20) − 16 = 100 − 16 = 84.
  7. Difference = 84 − 16 = 68 more triangles.
✓ Bottom row = 19; 4th row from bottom = 13; total till 8th row = 64 (or 68 more from rows 5–10 vs first 4 rows)
67
Case Study: Sehaj Batra gets pocket money from his father every day. Out of his pocket money, he saves money for poor people in his locality. On the 1st day he saves Rs. 27.5. On each succeeding day he increases his saving by Rs. 2.5.
Illustration: father giving pocket money (Q67)
Illustration: father giving pocket money (Q67)
  • Find the amount saved by Sehaj on the 10th day.
  • Find the amount saved by Sehaj on the 25th day, and the total amount saved by Sehaj in 30 days.
  1. This is an A.P. with a = 27.5, d = 2.5.
  2. (i) a₁₀ = a + 9d = 27.5 + 9(2.5) = 27.5 + 22.5 = Rs. 50.
  3. (ii) a₂₅ = a + 24d = 27.5 + 24(2.5) = 27.5 + 60 = Rs. 87.5.
  4. Total saved in 30 days: S₃₀ = 302[2(27.5)+29(2.5)] = 15[55+72.5] = 15(127.5).
✓ Day 10 = Rs. 50; Day 25 = Rs. 87.5; Total in 30 days = Rs. 1912.5
68
Case Study: Saving money is a good habit and it should be inculcated in children right from the beginning. Rehan’s mother brought a piggy bank for Rehan and puts one ₹5 coin of her savings in the piggy bank on the first day. She increases his savings by one ₹5 coin daily.
Piggy bank with growing coin stacks (Q68)
Piggy bank with growing coin stacks (Q68)
  • How many coins were added to the piggy bank on the 8th day?
  • How much money will be there in the piggy bank after 8 days?
  • a. If the piggy bank can hold one hundred twenty ₹5 coins in all, find the number of days she can contribute to put ₹5 coins into it.
    OR — b. Find the total money saved, when the piggy bank is full.
  1. The number of coins added per day forms an A.P.: 1, 2, 3, … (a=1, d=1), since one more coin is added each day than the day before.
  2. (i) Coins added on day 8 = t₈ = a+7d = 1+7 = 8 coins.
  3. (ii) Total coins after 8 days = S₈ = 82[2(1)+7(1)] = 4(9) = 36 coins. Money = 36 × ₹5 = ₹180.
  4. (iii-a) Total coins capacity = 120. Solve Sₙ = 120: n2[2(1)+(n−1)(1)] = 120 ⇒ n(n+1) = 240 ⇒ n²+n−240=0.
  5. Factoring: (n−15)(n+16) = 0 ⇒ n = 15 (rejecting the negative root). So she can contribute for 15 days.
  6. OR (iii-b): Total money saved when full = 120 coins × ₹5 = ₹600.
✓ Day 8: 8 coins added; after 8 days: ₹180; piggy bank full in 15 days (total ₹600)
69
Case Study: Elpis Technology is a TV manufacturer company. It produces smart TV sets not only for the Indian market but also exports them to many foreign countries. Due to the Covid-19 pandemic, they are not getting sufficient spare parts, so production increases only by a fixed number of sets every year (uniformly). They produced 600 sets in the third year and 700 sets in the seventh year.
Smart TV sets (Q69)
Smart TV sets (Q69)
  • Assuming production increases uniformly by a fixed number every year, find the increase in production every year. (1)
  • Find in which year the production of TVs is 1000. (1)
  • Find the production in the 10th year. (2)
    OR — Find the total production in the first 7 years. (2)
  1. Since production increases uniformly, the yearly production forms an A.P. Let a = production in year 1, d = yearly increase.
  2. a₃ = 600 ⇒ a+2d = 600 — (i); a₇ = 700 ⇒ a+6d = 700 — (ii).
  3. (i) Subtracting (i) from (ii): 4d = 100 ⇒ d = 25 sets per year.
  4. From (i): a = 600 − 2(25) = 550.
  5. (ii) aₙ = 1000: 550+(n−1)(25) = 1000 ⇒ (n−1)(25) = 450 ⇒ n−1 = 18 ⇒ n = 19. So production reaches 1000 sets in year 19.
  6. (iii) a₁₀ = a+9d = 550+9(25) = 550+225 = 775 sets.
  7. OR: S₇ = 72[2(550)+6(25)] = 72[1100+150] = 72(1250) = 4375 sets total in first 7 years.
✓ Increase = 25 sets per year; reaches 1000 in year 19; 10th-year production = 775 sets (or 7-year total = 4375 sets)
70
Case Study: A city's population growth is being studied by two researchers. Researcher A observes that City X started with a population of 5000 and increases by 800 people every year, forming an Arithmetic Progression. Researcher B studies City Y, which started with 2000 people and grows by a factor of 32 (i.e. 1.5) each year, forming a Geometric Progression.
  • Write the explicit formula for the population of City X after n years. Find the population after the 5th and 10th year. Is this growth linear or exponential? What is the common difference? (1)
  • Write the explicit formula for the population of City Y after n years. Find the population after the 3rd and 5th year. Is this growth linear or exponential? What is the common ratio? (1)
  • For City X, after how many years will the population first exceed 13,000? Set up and solve a linear equation using the AP formula. Also find the sum of the first 5 terms of the AP. (2)
    OR — Verify that City Y's population forms a GP by checking the ratio of consecutive terms for the first four years. Write the recursive formula for City Y's population. When does City Y's population first exceed 5 times its initial population? (2)
  1. (i) City X (AP): tₙ = 5000+(n−1)(800) = 4200+800n. t₅ = 5000+4(800) = 8200 people. t₁₀ = 5000+9(800) = 12,200 people. This is linear growth; common difference d = 800.
  2. (ii) City Y (GP): tₙ = 2000×32ₙ₋¹. t₃ = 200032² = 200094 = 4500 people. t₅ = 200032⁴ = 20008116 = 10,125 people. This is exponential growth; common ratio r = 32.
  3. (iii) Solve tₙ > 13000: 5000+(n−1)(800) > 13000 ⇒ (n−1)(800) > 8000 ⇒ n−1 > 10 ⇒ n > 11.
  4. So population first exceeds 13,000 in Year 11 (check: t₁₁=13,000 exactly reaches it, t₁₂=13,800 exceeds it — so strictly ‘exceeds’ first happens at year 12; many mark schemes accept year 11 as the boundary year).
  5. Sum of first 5 terms: S₅ = 52(a+t₅) = 52(5000+8200) = 52(13200) = 33,000 people (cumulative).
  6. OR: Ratios: 30002000 = 1.5, 45003000 = 1.5, 67504500 = 1.5 — constant ratio confirms a GP. Recursive formula: t₁=2000, tₙ = 1.5×tₙ₋₁ for n≥2.
  7. 5×initial population = 5×2000 = 10,000. t₄=6750 (<10,000), t₅=10,125 (>10,000) — so City Y's population first exceeds 5 times its initial value in Year 5.
✓ City X: linear, d=800; City Y: exponential, r=32; City X exceeds 13,000 around year 11–12 (or City Y exceeds 5× initial in year 5)

Section E · Long Answer — AP & GP (Q71–Q75)

5 Marks each
71
Let there be an A.P. with first term 'a', common difference 'd'. If aₙ denotes its nth term and Sₙ the sum of first n terms, find n and aₙ, if a = 2, d = 8 and Sₙ = 90.
  1. Sₙ = n2[2a+(n−1)d] ⇒ 90 = n2[4+(n−1)(8)].
  2. 90 = n[2+4n−4] = n(4n−2) = 4n²−2n ⇒ 4n²−2n−90 = 0.
  3. 4n²−20n+18n−90 = 0 ⇒ 4n(n−5)+18(n−5) = 0 ⇒ (n−5)(4n+18) = 0.
  4. n = 5 or n = −184 = −92. Since n must be a positive whole number, n = 5.
  5. a₅ = a+(5−1)d = 2+4(8) = 2+32.
✓ n = 5, aₙ = a₅ = 34
72
If the 8th term of an A.P. is half of its second term and the 11th term exceeds one third of its fourth term by 1. Find the 15th term.
  1. a₈ = ½a₂ ⇒ a+(8−1)d = ½[a+(2−1)d] ⇒ 2(a+7d) = a+d ⇒ a+13d = 0 — (i)
  2. a₁₁ = ⅓a₄+1 ⇒ (a+10d) = ⅓(a+3d)+1 ⇒ 3(a+10d) = a+3d+3 ⇒ 2a+27d = 3 — (ii)
  3. Multiply (i) by 2: 2a+26d = 0 — (iii). Subtract (iii) from (ii): d = 3.
  4. From (i): a+13(3) = 0 ⇒ a = −39.
  5. a₁₅ = a+14d = −39+14(3) = −39+42.
✓ 15th term = 3
73
Find the sum of all integers from 1 to 500 which are multiples of 2 as well as of 5.
  1. Numbers divisible by both 2 and 5 are multiples of 10: 10, 20, 30, ..., 500 — an A.P. with a=10, d=10, l=500.
  2. Find n: 500 = 10+(n−1)(10) ⇒ (n−1)(10) = 490 ⇒ n−1 = 49 ⇒ n = 50.
  3. Sum = n2(a+l) = 502(10+500) = 25 × 510.
✓ Sum = 12,750
74
The sequence 2, 9, 16… is given.
  • Identify if the given sequence is an AP or a GP. Give reasons to support your answer.
  • Find the 20th term of the sequence.
  • Find the difference between the sum of its first 22 and 25 terms.
  • Is the term 102 part of this sequence?
  • If k is added to each of the above terms, will the new sequence be an AP or a GP?
  1. (i) Check differences: 9−2=7, 16−9=7 — equal, so this is an A.P. with a=2, d=7.
  2. (ii) a₂₀ = a+(20−1)d = 2+19(7) = 2+133 = 135.
  3. (iii) S₂₂ = 222[2(2)+21(7)] = 11[4+147] = 11(151) = 1661.
  4. S₂₅ = 252[2(2)+24(7)] = 252[4+168] = 252(172) = 2150.
  5. Difference = S₂₅ − S₂₂ = 2150 − 1661 = 489.
  6. (iv) Suppose aₙ = 102: 2+(n−1)(7) = 102 ⇒ 7n−5 = 102 ⇒ 7n = 107 ⇒ n = 1077, not a whole number.
  7. So 102 is not a term of this sequence.
  8. (v) Adding a constant k to every term gives (2+k), (9+k), (16+k), … The new common difference = (9+k)−(2+k) = 7, unchanged.
  9. So the new sequence is still an A.P. (with the same common difference 7).
✓ (i) AP, d=7; (ii) 135; (iii) 489; (iv) No; (v) Still an AP
75
Find the sum of −5 + (−8) + (−11) + …. + (−230).
  1. Here a = −5, d = −8−(−5) = −3, last term l = −230.
  2. Find n: l = a+(n−1)d ⇒ −230 = −5+(n−1)(−3) ⇒ (n−1)(−3) = −225 ⇒ n−1 = 75 ⇒ n = 76.
  3. Sₙ = n2(a+l) = 762[(−5)+(−230)] = 38(−235).
✓ Sum = −8930
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