If a 1 + a 2 + a 3 + … + a n = 3(2 n+1 – 2), for every n ≥ 1, then a 11 equals

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6144 — Easy The given expression is the partial sum $S_n = a_1+\dots+a_n$. Any single term is recovered as $a_n = S_n - S_{n-1}$. Apply this with $n=11$. 1 Write the two partial sums. Using $S_n = 3(2^{n+1}-2)$. $$\begin{aligned} &S_{11} = 3(2^{12}-2)\ &S_{10} = 3(2^{11}-2) \end{aligned}$$ 2 Subtract. $a_{11}=S_{11}-S_{10}$; the constants $-2$ cancel. $$\begin{aligned} &a_{11} = 3(2^{12}-2) - 3(2^{11}-2) = 3\big(2^{12}-2^{11}\big) \end{aligned}$$ 3 Simplify. Factor $2^{11}$ from the bracket. $$\begin{

CAT 2019 Slot 1 — QA Question 12

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