If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

Question

For positive integers $ n $, if $ 4 a_n = \frac{n^2 + 5n + 6}{4} $ and $$ S_n = \sum_{k=1}^{n} \left( \frac{1}{a_k} \right), \text{ then the value of } 507 S_{2025} \text{ is:} $$

Answer 1

Given $S_7 = 49$ and $S_{17} = 289$, using the formula $S_n = \frac n2 [2a + (n-1)d]$

$S_7 = \frac 72 [2a + (7-1)d]$

$49 = \frac {7}{2} (2a + 6d)$

$7 = a + 3d$

$a + 3d = 7 ……..(i)$

Similarly,

$S_{17 }= \frac {17}{2} [2a + (17-1)d]$

$289 = \frac {17}{2} (2a + 16d)$

$17 = a + 8d$

$a + 8d = 17 …….(ii)$

Subtracting equation (i) from equation (ii):

$5d = 10$

$d = 2$

From equation (i):

$a + 3(2) = 7$

$a + 6 = 7$

$a = 1$

Using the formula $S_n = \frac n2 [2a + (n-1)d]$

$S_n = \frac n2 [2(1) + (n-1)(2)]$

$S_n = \frac n2 (2 + 2n - 2)$

$S_n = \frac n2 (2n)$

$S_n = n^2$

If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

Solution and Explanation

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