JEE Main 2022 (Online) 24th June Morning ShiftQuestion 1
If $\{a_i\}_{i=1}^{n}$, where $n$ is an even integer, is an arithmetic progression with common difference $1$, and $\sum_{i=1}^{n}a_i=192$, $\sum_{i=1}^{n/2}a_{2i}=120$, then $n$ is equal to :
Sequence and Series · Arithmetic Progression
Solution
Formulas used. (1) $S_n=\frac n2\left[2a+(n-1)d\right]$(2) $a_2,a_4,\dots$ is an A.P. with first term $a+1$ and difference $2$ (here $d=1$)Step 1 - All terms. $$\frac n2\left(2a+n-1\right)=192\ \Longrightarrow\ n(2a+n-1)=384 .$$ Step 2 - Even-indexed terms ($\tfrac n2$ terms): $$\frac n4\left[2(a+1)+\left(\frac n2-1\right)2\right]=\frac n4(2a+n)=120\ \Longrightarrow\ n(2a+n)=480 .$$ Step 3 - Subtract. $$n(2a+n)-n(2a+n-1)=n=480-384=96 .$$ Answer: (B)