CAT 2019 Slot 1 QA Question & Solution
AlgebraMedium
Question
If $a_1 + a_2 + a_3 + .... + a_n = 3(2^{n + 1} - 2)$, for every $n \geq 1$, then $a_{11}$ equals
Solution
11th term of series = $a_{11}$ = Sum of 11 terms - Sum of 10 terms = $3(2^{11 + 1} - 2)$-3$(2^{10 + 1} - 2)$
= 3$(2^{12} - 2-2^{11} +2)$=3$(2^{11})(2-1)$= 3*$2^{11}$ = 6144