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CAT 2019 Quant Question with Solution 23

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

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Correct Answer: 6144

If n =1, ${{a}_{1}}=3\left( {{2}^{1+1}}-2 \right)=6=3\times {{2}^{1}}$

If n=2, ${{a}_{1}}+{{a}_{2}}=3\left( {{2}^{2+1}}-2 \right)=18$$\Rightarrow {{a}_{2}}=18-{{a}_{1}}=12=3\times {{2}^{2}}$

If n=3,  ${{a}_{1}}+{{a}_{2}}++{{a}_{3}}=3\left( {{2}^{3+1}}-2 \right)=42$$\Rightarrow {{a}_{3}}=42-\left( {{a}_{1}}+{{a}_{2}} \right)=24=3\times {{2}^{3}}$

Following the pattern, ${{a}_{n}}=3\times {{2}^{n}}$

Therefore, ${{a}_{11}}=3\times {{2}^{11}}=6144$

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