Class 11 Maths Chapter 8: Sequences and Series MCQs with answers
10 multiple-choice questions on Sequences and Series, each with its answer and the reason, written from the NCERT chapter. For school exams on the CBSE pattern and JEE Main, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.
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Answer: (a) 3
a7 = (−1)7(19) = −19 and a8 = (+1)(22) = 22, so a7 + a8 = 3.
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Answer: (b) 32
The terms are 1, 2, 2, 4, 8, 32: a3 = 2 × 1, a4 = 2 × 2, a5 = 4 × 2, a6 = 8 × 4 = 32.
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Answer: (c) 3, −6, 12, −24, …
In 3, −6, 12, −24 each term is non-zero and each ratio is −2. 1, 4, 9, 16 has ratios 4, 9/4, 16/9; 2, 4, 6, 8 has a constant difference, not ratio; a G.P. needs every term non-zero, so 1, 0, 0, 0 is not one.
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Answer: (d) 23
a = 486, r = −162/486 = −1/3, so a7 = 486 × (−1/3)6 = 486/729 = 2/3.
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Answer: (b) 3
−1 + 2 − 3 + 4 − 5 + 6 = 3.
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Answer: (a) 8532
a = 4, r = −1/2: S8 = 4[1 − (−1/2)8]/(1 + 1/2) = 4 × (255/256) × (2/3) = 85/32.
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Answer: (d) 11th term
an = (5/4)2n − 1 > 1000 needs 2n − 1 > 800. Since 29 = 512 and 210 = 1024, n − 1 = 10 and n = 11 (the term is 1280).
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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.
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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.
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Answer: (d) k = 2 or k = −3
(k + 1)2 = (k − 1)(2k + 5) gives k2 + 2k + 1 = 2k2 + 3k − 5, i.e. k2 + k − 6 = 0, so k = 2 or −3. k = 2 gives 1, 3, 9 and k = −3 gives −4, −2, −1; both are G.P.s (ratios 3 and 1/2).