The sum of the first 15 terms of the A.P. -10 | vedclass

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Question

(C) The given $A.P.$ is $-10, -12, -14, -16, \ldots$Here,the first term $a = -10$ and the common difference $d = -12 - (-10) = -12 + 10 = -2$.The numb

Vedclass · Accepted Answer

$-360$

Vedclass · Answer

(C) The given $A.P.$ is $-10, -12, -14, -16, \ldots$Here,the first term $a = -10$ and the common difference $d = -12 - (-10) = -12 + 10 = -2$.The number of terms $n = 15$.The formula for the sum of the first $n$ terms of an $A.P.$ is $S_n = \frac{n}{2} [2a + (n - 1)d]$.Substituting the values: $S_{15} = \frac{15}{2} [2(-10) + (15 - 1)(-2)]$.$S_{15} = \frac{15}{2} [-20 + (14)(-2)]$.$S_{15} = \frac{15}{2} [-20 - 28]$.$S_{15} = \frac{15}{2} [-48]$.$S_{15} = 15 	imes (-24) = -360$.Therefore,the sum of the first $15$ terms is $-360$.

The sum of the first $15$ terms of the $A.P.$ $-10, -12, -14, -16, \ldots$ is:

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