Question

The sum of the first three terms of an arithmetic progression is $9$ and the sum of their squares is $35.$ The sum of the first $n$ terms of the series can be

• A. $n\left(n + 1\right)$
• B. $2n^{2}$
• C. $n\left(4 - n\right)$
• D. $n\left(6 - n\right)$

Answer 1

Answer: (D)

Let $a-d,a\&a+d$ are three numbers in A.P.
Given, $\left(a - d\right)+\left(a\right)+\left(a + d\right)=9\Rightarrow a=3$
and $\left(a - d\right)^{2}+a^{2}+\left(a + d\right)^{2}=35$
$\Rightarrow 3a^{2}+2d^{2}=35\Rightarrow d=\pm2$
$a=3,d=2\Rightarrow 1,3,5,\ldots \ldots ..$
$S_{n}=\frac{n}{2}\left(2 + \left(n - 1\right) 2\right)=n^{2}$
$a=3,d=-2\Rightarrow 5,3,1,\ldots \ldots ..$
$S_{n}=\frac{n}{2}\left[10 + \left(n - 1\right) \left(- 2\right)\right]=n\left(5 - n + 1\right)$
$=n\left(6 - n\right)$