#### The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167. If the sum of the first ten terms of this AP is 235, find the sum of its first twenty terms.

$\\S\textsubscript{5} + S\textsubscript{7} = 167 \ldots (1)\\ S\textsubscript{10} = 235 \ldots (2)\\$

We know that

$S_{n} = \frac{n}{2}\left( 2a + (n-1)d \right) \\ \text{Put } n = 5, S_{5} = \frac{5}{2}\left( 2a + (5-1)d \right) \\ S_{5} = \frac{5}{2}\left( 2a + 4d \right) \\ \text{Put } n = 7, S_{7} = \frac{7}{2}\left( 2a + (7-1)d \right) \\ S_{7} = \frac{7}{2}\left( 2a + 6d \right) \\ \text{Put } n = 10, S_{10} = \frac{10}{2}\left( 2a + (10-1)d \right) \\ S_{10} = 5(2a + 9d) \\ \text{Now put the values of } S_{5}, S_{7} \text{ and } S_{10} \text{ in equations (1) and (2) we get:} \\ \frac{5}{2}\left( 2a + 4d \right) + \frac{7}{2}\left( 2a + 6d \right) = 167 \\ 5a + 10d + 7a + 21d = 167 \\ 12a + 31d = 167 \ldots (3) \\ 5(2a + 9d) = 235 \\ 2a + 9d = \frac{235}{5} \\ 2a + 9d = 47 \ldots (4) \\ \text{Solving equations (3) and (4):} \\ 12a + 31d = 167 \ldots (3) \\ 12a + 54d = 282 \quad \{ \text{Multiply (4) by 6} \}$
$\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ -23d = - 115\\ d=\frac{-115}{-23} \\ d = 5\\ Put d = 5 in (3) we get\\ 12a + 31(5) = 167 \\ 12a = 167 - 155 \\ 12a = 12\\ a=\frac{12}{12}=1 \\ = 10 [2 + 95]\\ S\textsubscript{20} = 970\\$