In an AP, if $S_{5} + S_{7} = 167$ and $S_{10} = 235$ , then find the A.P., where $S_{n}$ denotes the sum of its first $n$ terms.

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Solution

Let the first term is $a$ and the common difference is $d$

By using $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ we have,

$S_{5} = \frac{5}{2} [2 a + (5 - 1) d] = \frac{5}{2} [2 a + 4 d]$

$S_{7} = \frac{7}{2} [2 a + (7 - 1) d] = \frac{7}{2} [2 a + 6 d]$

Given: $S_{7} + S_{5} = 167$

$∴ \frac{5}{2} [2 a + 4 d] + \frac{7}{2} [2 a + 6 d] = 167$

$\Rightarrow 10 a + 20 d + 14 a + 42 d = 334$

$\Rightarrow 24 a + 62 d = 334$ ...(1)

$S_{10} = \frac{10}{2} [2 a + (10 - 1) d]$ = $5 (2 a + 9 d)$

Given: $S_{10} = 235$

So $5 (2 a + 9 d) = 235$

$\Rightarrow 2 a + 9 d = 47$ ...(2)

Multiply equation (2) by $12$ , we get

$24 a + 108 d = 564 . . . . (3)$

Subtracting equation (3) from (1), we get

$- 46 d = - 230$

$∴ d = 5$

Substing the value of $d = 5$ in equation (1) we get

$2 a + 9 (5) = 47$ or $2 a = 2$

$∴ a = 1$

Then A.P is $1, 6, 11, 16, 21, \dots$

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