If $a_1,\ a_2,\ a_3,\ ....$ are in A.P. such that $a_1+a_7+a_{16}=40$, then find the sum of the first $15$ terms of this A.P.

Given: $a_1,\ a_2,\ a_3,\ ....$ are in A.P. such that $a_1+a_7+a_{16}=40$.

To do: To find the sum of the first $15$ terms of this A.P.

Solution:

$a_1+a_7+a_{16}=40$

$\Rightarrow a+a+6d+a+15d=40$

$\Rightarrow 3a+21d=40$

$\Rightarrow a+7d=\frac{40}{3}$

​

$S_{15}=\frac{15}{2}( 2a+14d)=15( a+7d)$

$S_{15}=15\times\frac{40}{3}$

​

$=200$

$\Rightarrow S_{15}=200$.

Thus, sum of $15$ terms of the given A.P. is $200$.

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Updated on: 10-Oct-2022

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