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Finding a Specific Term from Sum of Series

CAT 2019 Slot 1 · Quantitative Ability · Medium · Progressions

This is a medium Quantitative Ability question from the CAT 2019 Slot 1 paper. It tests Progressions. 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 a1+a2+a3+⋯+an=3(2n+1−2)a_1 + a_2 + a_3 + \dots + a_n = 3(2^{n+1} - 2), for every n≥1n \ge 1, then a11a_{11} equals

Answer

6144

Explanation

Sn=3(2n+1−2)S_n = 3(2^{n+1} - 2). a11=S11−S10=3(212−2)−3(211−2)=3(212−211)=3⋅211=3⋅2048=6144a_{11} = S_{11} - S_{10} = 3(2^{12} - 2) - 3(2^{11} - 2) = 3(2^{12} - 2^{11}) = 3 \cdot 2^{11} = 3 \cdot 2048 = 6144.

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