# ISI2014-DCG-62

0 votes
65 views

If the sum of the first $n$ terms of an arithmetic progression is $cn^2$, then the sum of squares of these $n$ terms is

1. $\frac{n(4n^2-1)c^2}{6}$
2. $\frac{n(4n^2+1)c^2}{3}$
3. $\frac{n(4n^2-1)c^2}{3}$
4. $\frac{n(4n^2+1)c^2}{6}$

recategorized

## 1 Answer

1 vote

Answer: $\mathbf C$

Explanation:

$\mathrm {S_n = cn^2} \text{[Given]}$

Now $\mathrm {n^{th}}$ term from the sum can be calculated by using, the formula:

$\mathrm {\therefore t_n = S_n - S_{n-1} = cn^2-c(n-1)^2 = c(2n-1)}$

Now,

$\mathrm {\sum t_{n^2} = \sum c^2(2n-1)^2 = \sum c^2(4n^2+1-4n)}\tag{i}$

Now, we know that:

$\mathrm {\sum n^2 = \frac{n(n+1)(2n+1)}{6}\;,\; \sum n = \frac{n(n+1)}{2}\;, and\;\sum 1 = n}$

Substituting these values in $\mathrm {(i)}$, we get:

$\mathrm {\sum t_{n^2} = c^2\Big(4\frac{n(n+1)(2n+1)}{6}\;+n + \frac{n(n+1)}{2}\Big) = \frac{c^2(4n^3-n)}{3} = \frac{n(4n^2-1)c^2}{3}}$

$\therefore \mathbf C$ is the right option.

edited by

## Related questions

2 votes
2 answers
1
153 views
The sum of the series $\:3+11+\dots +(8n-5)\:$ is $4n^2-n$ $8n^2+3n$ $4n^2+4n-5$ $4n^2+2$
0 votes
1 answer
2
122 views
If $l=1+a+a^2+ \dots$, $m=1+b+b^2+ \dots$, and $n=1+c+c^2+ \dots$, where $\mid a \mid <1, \: \mid b \mid < 1, \: \mid c \mid <1$ and $a,b,c$ are in arithmetic progression, then $l, m, n$ are in arithmetic progression geometric progression harmonic progression none of these
4 votes
3 answers
3
322 views
The number of divisors of $6000$, where $1$ and $6000$ are also considered as divisors of $6000$ is $40$ $50$ $60$ $30$
1 vote
2 answers
4
179 views
The sum of the series $\dfrac{1}{1.2} + \dfrac{1}{2.3}+ \cdots + \dfrac{1}{n(n+1)} + \cdots$ is $1$ $1/2$ $0$ non-existent