Find the sum of first 16 terms of the A.P. 10, 6, 2, ............ .

An A.P. 10, 6, 2, ............ is given and we need to find the sum of first 16 terms.

The common difference of A.P. is 6-10=-4

It is known that the sum of an A.P. is given by

${\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2\mathrm{a}+(\mathrm{n}-1\left)\mathrm{d}\right]$, where Sn is the sum of terms, n is the number of terms, a is the first term and d is the common difference of A.P.

On substituting the values, we get the sum as

$\begin{array}{l}\Rightarrow {\mathrm{S}}_{16}=\frac{16}{2}\left[2\right(10)+(16-1\left)\right(-4\left)\right]\\ \Rightarrow {\mathrm{S}}_{16}=8[20+(15\left)\right(-4\left)\right]\\ \Rightarrow {\mathrm{S}}_{16}=8[20+(-60\left)\right]\\ \Rightarrow {\mathrm{S}}_{16}=8[-40]\\ \Rightarrow {\mathrm{S}}_{16}=-320\end{array}$

 Hence, the sum of first 16 terms of the A.P. is -320.