CAT 2019 Quant Question with Solution 23

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