Let respectively be the sum to 12 terms of 10 A.P. whose first terms are 1, 2, 3, ....... 10 and the common difference are 1, 3, 5, .........., 19 respectively. Then is equal to [2023]

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Solution

Q.
Let $S_{1}, S_{2}, S_{3}, . . . . . . . . . . . . S_{10}$ respectively be the sum to 12 terms of 10 A.P. whose first terms are 1, 2, 3, ....... 10 and the common difference are 1, 3, 5, .........., 19 respectively. Then $\sum_{i = 1}^{10} S_{i}$ is equal to [2023]

1 7220

2 7380

3 7260

4 7360

Ans.
(3)

$s_{1} = 1 + 2 + 3 + \dots + 12 \dots (i)$

$s_{2} = 2 + 5 + 8 + \dots up to 12 terms \dots (i i)$

$s_{3} = 3 + 8 + 13 + \dots up to 12 terms \dots (i i i)$

$∵$ $s_{10} = 10 + 29 + 48 + \dots up to 12 terms \dots (i v)$

$From (i), s_{1} = \frac{12 (13)}{2} = 78$

$From (ii), s_{2} = \frac{12}{2} [2 (2) + 11 \times 3] = 6 [4 + 33] = 222$

$From (iii), s_{3} = \frac{12}{2} [2 (3) + 11 \times 5] = 6 [6 + 55] = 366$

$From (iv), s_{10} = \frac{12}{2} [2 (10) + 11 \times 19] = 6 [20 + 209] = 1374$

$Thus, s_{k} = 6 [2 k + 11 (2 k - 1)] = 6 (2 k + 22 k - 11) = 144 k - 66$

$∴$ $\sum_{k = 1}^{10} s_{k} = 144 \sum_{k = 1}^{10} k - 66 \sum_{k = 1}^{10} 1 = 144 (\frac{10 \times 11}{2}) - 66 (10)$

$= 7920 - 660 = 7260$

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