#### If sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.

$\\S\textsubscript{6} = 36, S\textsubscript{16} = 256\\ \left[ Using\,\,{{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \right] \\ S\textsubscript{6} = 36 S\textsubscript{16} = 256\\ = \frac{6}{2}\left[ 2a+\left( 6-1 \right)d \right]=36 \frac{16}{2}\left[ 2a+\left( 16-1 \right)d \right]=256 \\ =3[2a + 5d] = 36 8[2a + 15d] = 256\\ =2a + 5d = 12 \ldots (1) \\2a + 15d = 32 \ldots (2)\\\text{from (1)and (2)}$

$\\ \_\_\_\_\_\_\_\_\_\_\_\_\\ 10d = 20\\ d = 2\\ Put \ d = 2\ in (1)\\ 2a + 5(2) = 12\\ 2a = 12 - 10\\ a = 1\\ ( \left[ Using\,\,{{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \right] \\ Put n= 1\\ = 5 [2 + 18] \\ 5[20]\\ S\textsubscript{10} = 100\\$