Question:
If $a_{1}, a_{2}, a_{3}, \ldots .$ are in A.P. such that $a_{1}+a_{7}+a_{16}=40$, then the sum of the first 15 terms of this A.P. is :
Correct Option: 1
Solution:
$a_{1}+a_{7}+a_{16}=40$
$a+a+6 d+a+15 d=40$
$\Rightarrow 3 \mathrm{a}+21 \mathrm{~d}=40 \quad \Rightarrow \mathrm{a}+7 \mathrm{~d}=\frac{40}{3}$
$\mathrm{S}_{15}=\frac{15}{2}(2 \mathrm{a}+14 \mathrm{~d})=15(\mathrm{a}+7 \mathrm{~d})$
$\mathrm{S}_{15}=15 \times \frac{40}{3} \Rightarrow 200 \quad \mathrm{~S}_{15}=200$