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# 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 $

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 $-320$.

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