JEE Main 2023 (Online) 13th April Morning ShiftQuestion 1
Let $s_1,s_2,s_3,\ldots,s_{10}$ respectively be the sum to $12$ terms of $10$ A.P.s whose first terms are $1,2,3,\ldots,10$ and the common differences are $1,3,5,\ldots,19$ respectively. Then $\sum_{i=1}^{10}s_i$ is equal to
Sequence and Series · Arithmetic Progression
Solution
Formulas used. (1) $S_n=\frac n2\left[2a+(n-1)d\right]$Step 1 - The $i$-th A.P. First term $i$, common difference $2i-1$, $12$ terms: $$s_i=6\left[2i+11(2i-1)\right]=6(24i-11)=144i-66 .$$ Step 2 - Add over $i=1$ to $10$. $$\sum_{i=1}^{10}s_i=144(55)-66(10)=7920-660=7260 .$$ Answer: (A)