CAT 2019 Set-1 | Question: 89

Question

If $a_{1}+a_{2}+a_{3}+\dots+a_{n}=3(2^{n+1}-2)$, for every $n\geq 1$, then $a_{11}$ equals ______

Answer 1

Given that, $ a_{1} + a_{2} + a_{3} + \dots + a_{n} = 3(2^{n+1} – 2) $

Now, put $ n=11,$ we get

$ a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}+a_{8}+a_{9}+a_{10}+a_{11} = 3(2^{12}-2) $

$ = 3 (4096-2) $

$ = 3 \times 4094 $

$ = 12282 \quad \longrightarrow (1) $

And put $ n=10,$ we get

$ a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}+a_{8}+a_{9}+a_{10} = 3(2^{11}-2) $

$ = 3 (2048 – 2) $

$ = 3 \times 2046 $

$ = 6138 \quad \longrightarrow (2) $

From  the equation $(1),$ subtract the equation $(2),$ we get

$ a_{11} = 12282 – 6138 $

$ \boxed{a_{11} = 6144} $

Correct Answer $: 6144$