Find the sum of first $16$ terms of the AP: $10,\ 6,\ 2\ .......$
Given: AP: $10,\ 6,\ 2\ .......$
To do: To find the sum of first $16$ terms of the A.P.
Solution:
$a=10,\ n=16,\ d=6-10=-4$
We know that
$S_n=\frac{n}{2}[2a+( n-1)d]$
$S_{16}=\frac{16}{2}[2\times 10+(16-1)( -4)]$
$\Rightarrow S_{16}=8[20+15( -4)]$
$\Rightarrow S_{16}=8\times ( -40)$
$\Rightarrow S_{16}=-320$
Thus, the sum of first $16$ terms of the given A.P. is $-320$.
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