Question

If $a_1,a_2,a_3,..........,a_n$ are consecutive terms of an increasing $A.P.$ and $(1^2-a_1)+(2^2-a_2)+(3^2-a_3)+.......+(n^2-a_n)=\frac{(n-1)n(n+1)}{3}$, then the value of $\left(\frac{a_5+a_3-a_2}{6}\right)$ is equal to

• A. $1$
• B. $-1$
• C. $2$
• D. $-2$

Answer 1

The correct option is A $1$

Given, $a_1,a_2...a_n$ are in A.P

Also, $(1^2-a_1)+(2^2-a_2)+...+(n^2-a_n)=\frac{(n-1)n(n+1)}{3}...\left(1\right)$

$\Rightarrow (1^2+2^2+...n^2)-(a_1+a_2+...+a_n)=\frac{(n-1)n(n+1)}{3}$

$\sum n^2-\frac{n}{2}(a_1+a_n)=\frac{n(n-1)(n+1)}{3}$ [ sum of AP]

$\frac{n(n+1)(2n+1)}{6}-\frac{n(n-1)(n+1)}{3}=\frac{n}{2}(a_1+a_n)[\sum n^2$ formula]

$\frac{n+1}{3}[\frac{2n+1}{2}-(n-1\left)\right]=\frac{a_1+a_n}{2}$

$n+1=a_1+a_n...\left(2\right)$

putting n =1 in (1)$\Rightarrow 1^2-a_1=0\Rightarrow a_1=1$

from(2), $a_n=n$

$\therefore \left(\frac{a_5+a_3-a_2}{6}\right)=\left(\frac{a_5+d}{6}\right)=\frac{a_6}{6}=\frac{6}{6}=1$ [$\because a_3-a_2=d]$ common diff.

option A is correct