Class 11 Maths Chapter 8 Sequences and Series Worksheet with Answers, free PDF (CBSE Class 11 and JEE Main; based on NCERT)

A 20-mark worksheet on Sequences and Series for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.

Question 1

For the sequence a_n = (−1)ⁿ(3n − 2), the value of a₇ + a₈ is

(a) 3(b) −3(c) 41(d) −41

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Answer: (a) 3

a₇ = (−1)⁷(19) = −19 and a₈ = (+1)(22) = 22, so a₇ + a₈ = 3.

Question 2

The 2nd term of a G.P. is 6 and its 5th term is −48. The sum of its first four terms is

(a) −15(b) 45(c) 15(d) −45

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Answer: (c) 15

ar⁴/ar = r³ = −48/6 = −8, so r = −2 and a = 6/(−2) = −3. The terms are −3, 6, −12, 24, with sum 15.

Question 3

The value of Σ_k=1⁶ (−1)^k k is

(a) −3(b) 3(c) 21(d) 0

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Answer: (b) 3

−1 + 2 − 3 + 4 − 5 + 6 = 3.

Question 4

Three numbers are inserted between 2 and 162 so that the five numbers form a G.P. The product of the three inserted numbers is

(a) 324(b) 1944(c) 5832(d) 18

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Answer: (c) 5832

r⁴ = 162/2 = 81, so r = ±3. The product G₁G₂G₃ = (2r)(2r²)(2r³) = 8r⁶ = 8 × 729 = 5832 for either sign; it is also (√(2 × 162))³ = 18³.

Question 5

The 7th term of the G.P. 486, −162, 54, … is

(a) −23(b) −2(c) 6(d) 23

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Answer: (d) 23

a = 486, r = −162/486 = −1/3, so a₇ = 486 × (−1/3)⁶ = 486/729 = 2/3.

Question 6 · 2 marks

A culture of 500 cells triples every 2 hours. After how many hours will it have 40,500 cells?

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After k two-hour periods there are 500 × 3^k cells. 500 × 3^k = 40500 gives 3^k = 81 = 3⁴, so k = 4 periods, which is 8 hours.

500 × 3^k = 40500 gives k = 4 (1)
8 hours (1)

Question 7 · 2 marks

Show that the sequence a_n = 5 × 3^{2n − 1} is a G.P., and write its first term and common ratio.

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a_{n + 1}/a_n = 5 × 3^{2n + 1}/(5 × 3^{2n − 1}) = 3² = 9, a constant, and every term is non-zero, so the sequence is a G.P. a₁ = 5 × 3 = 15 and r = 9.

a_{n + 1}/a_n = 9, constant: a G.P. (1)
first term 15, common ratio 9 (1)

Question 8 · 2 marks

The 4th term of an A.P. is 14 and its 9th term is 34. Find the sum of its first 12 terms.

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a + 3d = 14 and a + 8d = 34 give d = 4, a = 2. S₁₂ = (12/2)[2(2) + 11(4)] = 6 × 48 = 288.

a = 2, d = 4 (1)
S₁₂ = 288 (1)

Question 9 · 3 marks

The first term of a G.P. is 1, its last term is 128 and the sum of all its terms is 255. Find the common ratio and the number of terms.

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With l = ar^{n − 1}, S = (lr − a)/(r − 1). So 255 = (128r − 1)/(r − 1), 255r − 255 = 128r − 1, 127r = 254 and r = 2. Then 1 × 2^{n − 1} = 128 = 2⁷, so n = 8.

S = (lr − a)/(r − 1) written from S_n and l = ar^{n − 1} (1)
r = 2 (1)
n = 8 (1)

Question 10 · 3 marks

Meera saves ₹50 in the first week and, in every later week, ₹15 more than in the week before. How much has she saved in all at the end of 20 weeks?

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The weekly savings form an A.P. with a = 50 and d = 15. S₂₀ = (20/2)[2(50) + 19(15)] = 10 × (100 + 285) = ₹3850.

an A.P. with a = 50, d = 15 (1)
S₂₀ = 10[100 + 285] (1)
₹3850 (1)

Question 11 · 3 marks

Two positive numbers are in the ratio 1 : 4, and their A.M. exceeds their G.M. by 2. Find the numbers.

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Let the numbers be a and 4a (a > 0). A = 5a/2 and G = √(4a²) = 2a. A − G = a/2 = 2, so a = 4. The numbers are 4 and 16 (check: A = 10, G = 8).

A = 5a/2 and G = 2a (1)
a/2 = 2 (1)
the numbers 4 and 16 (1)

Class 11 Maths Chapter 8: Sequences and Series worksheet with answer key