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 ${\left(a-d\right)}^2+a^2+{\left(a+d\right)}^2=35$
$\Rightarrow 3a^2+2d^2=35\Rightarrow d=\pm 2$
$a=3,d=2\Rightarrow 1,3,5,\dots \dots ..$
$S_n=\frac{n}{2}\left(2+\left(n-1\right)2\right)=n^2$
$a=3,d=-2\Rightarrow 5,3,1,\dots \dots ..$
$S_n=\frac{n}{2}\left[10+\left(n-1\right)\left(-2\right)\right]=n\left(5-n+1\right)$
$=n\left(6-n\right)$