JEE Maths›Sequence & Series›Visual Solution
Visual SolutionPYQ 2023 · Apr 13 Shift 1Standard
Question
Let s1,s2,…,s10s_1, s_2, \ldots, s_{10} respectively be the sum of 12 terms of 10 A.P.s whose first terms are 1,2,3,…,101, 2, 3, \ldots, 10 and the common differences are 1,3,5,…,191, 3, 5, \ldots, 19 respectively. Then ∑i=110si\sum_{i=1}^{10} s_i is equal to
(A)72207220
(B)73607360
(C)72607260
(D)73807380
Solution Path
Each si=144i−66s_i = 144i - 66. Summing: 144⋅55−660=7260144 \cdot 55 - 660 = 7260.
01Question Setup
1/4
10 APs with first terms 1,2,…,101, 2, \ldots, 10 and CDs 1,3,5,…,191, 3, 5, \ldots, 19. Find ∑si\sum s_i.
∑i=110si=  ?\sum_{i=1}^{10} s_i = \;?
02Express s_i
2/4
For the ii-th AP: ai=ia_i = i, di=2i−1d_i = 2i - 1. Using AP sum formula: si=144i−66s_i = 144i - 66.
si=144i−66s_i = 144i - 66
03Sum AllKEY INSIGHT
3/4
∑si=144⋅10⋅112−66⋅10=7920−660=7260\sum s_i = 144 \cdot \dfrac{10 \cdot 11}{2} - 66 \cdot 10 = 7920 - 660 = 7260.
7920−660=72607920 - 660 = 7260
04Final Answer
4/4
∑i=110si=7260\sum_{i=1}^{10} s_i = 7260.
7260\boxed{7260}
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