If the sum of first n terms of an A.P. is c n | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$S _{ n }= cn ^2$

$S_{ n -1}= c ( n -1)^2= cn ^2+ c -2 cn$

$T _{ n }=2 cn - c$

$T _{ n }^2=(2 cn - c )^2=4 c ^2 n ^2+ c ^2-4 c ^2 n$

Requi

Vedclass · Accepted Answer

$\frac{n\left(4 n^2-1\right) c^2}{3}$

Vedclass · Answer

$S _{ n }= cn ^2$

$S_{ n -1}= c ( n -1)^2= cn ^2+ c -2 cn$

$T _{ n }=2 cn - c$

$T _{ n }^2=(2 cn - c )^2=4 c ^2 n ^2+ c ^2-4 c ^2 n$

Required sum

$=\sum T_n^2=4 c^2 \sum n^2+n c^2-4 c^2 \sum n$

$=\frac{4 c^2 n(n+1)(2 n+1)}{6}+n c^2-2 c^2 n(n+1)$

$=\frac{2 c^2 n(n+1)(2 n+1)+3 n c^2-6 c^2 n(n+1)}{3}$

$=\frac{n c^2\left[4 n^2+6 n+2+3-6 n-6\right]}{3}$

$=\frac{n c^2\left(4 n^2-1\right)}{3}$

If the sum of first $n$ terms of an $A.P.$ is $c n^2$, then the sum of squares of these $n$ terms is

Similar Questions

If the sum and product of the first three term in an $A.P$. are $33$ and $1155$, respectively, then a value of its $11^{th}$ tern is

The number of terms of the $A.P. 3,7,11,15...$ to be taken so that the sum is $406$ is

Suppose $a_{1}, a_{2}, \ldots, a_{ n }, \ldots$ be an arithmetic progression of natural numbers. If the ratio of the sum of the first five terms of the sum of first nine terms of the progression is $5: 17$ and $110< a_{15} < 120$ , then the sum of the first ten terms of the progression is equal to -

If the ${n^{th}}$ term of an $A.P.$ be $(2n - 1)$, then the sum of its first $n$ terms will be

Let ${S_n}$ denotes the sum of $n$ terms of an $A.P.$ If ${S_{2n}} = 3{S_n}$, then ratio $\frac{{{S_{3n}}}}{{{S_n}}} = $