Recursive Sequence Term

CAT 2000 Slot 1 · Quantitative Ability · Medium · Sequences and Series

This is a medium Quantitative Ability question from the CAT 2000 Slot 1 paper. It tests Sequences and Series. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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If $a_1 = 1$ and $a_{n+1} = 2a_n + 5$ for $n = 1, 2, \dots$ , then $a_{100}$ is equal to:

A.
$(5 \times 2^{99} - 6)$
B.
$(5 \times 2^{99} + 6)$
C.
$(6 \times 2^{99} + 5)$
D.
$(6 \times 2^{99} - 5)$

Answer

Explanation

$a_1 = 1$ $a_2 = 2(1) + 5 = 7$ $a_3 = 2(7) + 5 = 19$ $a_4 = 2(19) + 5 = 43$

The differences are $6, 12, 24, \dots$ which is a geometric progression with first term $6$ and common ratio $2$ . $a_n = a_1 + \sum_{k=1}^{n-1} 6 \cdot 2^{k-1} = 1 + 6(2^{n-1} - 1) = 6 \cdot 2^{n-1} - 5$ . For $n = 100$ , $a_{100} = 6 \times 2^{99} - 5$ .

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Recursive Sequence Term

Explanation

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