JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Sequences and Series

The AP is $3,8,13,\dots ,373$ with first term $a=3$ and common difference $d=5$. The number of terms is $n=\frac{373-3}{5}+1=75$.

Sum of all terms: $S_{\text{all}}=\frac{75}{2}(3+373)=75\times 188=14100$.

Terms divisible by 3 form the sub-sequence $3,18,33,\dots ,363$ (every 3rd term starting from the 1st). This is an AP with first term 3, common difference 15, and last term 363. Number of terms: $\frac{363-3}{15}+1=25$.

Sum of terms divisible by 3: $S_{\text{div3}}=\frac{25}{2}(3+363)=25\times 183=4575$.

Required sum: $14100-4575=9525$.