# Choose the correct answer from the given four options:The sum of first 16 terms of the AP: $10,6,2, \ldots$ is(A) $-320$(B) 320(C) $-352$(D) $-400$

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Given:

Given AP is $10,6,2, \ldots$.

To do:

We have to find the sum of the first 16 terms.

Solution:

$a_1 = a= 10, a_2=6$

Common difference $d =a_2-a_1$

$=6-10$

$=-4$

We know that,

Sum of the $n$ terms of an AP is $S_n=\frac{n}{2}[2a+(n-1)d]$

Therefore,

$S_{16}= \frac{16}{2}[2(10)+(16-1)(-4)]$

$= 8(20+15(-4))$

$= 8(20-60)$

$=8\times (-40)$

$=-320$

The sum of the first 16 terms of the AP is $-320$.

Updated on 10-Oct-2022 13:27:27