1. ## Arithmetic Progression

I need help with this question involving an A.P.

The question says: If the sum of $n$ terms of an A.P is $cn^2$ then the sum of squares of these $n$ terms will be equal to:

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

What I did: Let the sum of n terms of the A.P be $S_{n}$.

$\Rightarrow S_{n}=\frac{n}{2}[a_{1}+a_{n}]$

$\Rightarrow cn^2=\frac{n}{2}[a_{1}+a_{n}]$

$\Rightarrow 2cn=a_{1}+a_{n}$ ...(i)

When n=1

$\Rightarrow 2c=a_{1}+a_{1}$

$\Rightarrow c=a_{1} ...(ii)$

In equation (i)

$\Rightarrow 2cn-c=a_{n}$ Using (ii)

I have found the value of $a_{n}$. What should I do next?

2. ## Re: Arithmetic Progression

Seems no clue to square the terms just with $a_{1}$ and $a_{n}$
I would use difference method:
$a_{n}=S_{n}-S_{n-1}=c n^{2}-c(n-1)^{2}$

$c n^{2}-c(n-1)^{2}$ <-- square this and sum from 1 to n

$\sum_{r=1}^{n}[c r^{2}-c(r-1)^{2}]^{2}$ ...works... -->ans (c)

I'm trying to get used to TEX

3. ## Re: Arithmetic Progression

Thanks, I got it.

$a_{n}=2cn-c$

$\Rightarrow a_{n}^2=4c^2n^2+c^2-4c^2n$

I just have to find : $\sum_{n=1}^{n}a_{n}^2$

$\sum_{n=1}^{n}a_{n}^2$

$= \sum_{n=1}^{n}(4c^2n^2+c^2-4c^2n)$

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

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

So the correct answer is (c).

Am I correct?

4. ## Re: Arithmetic Progression

Sure. My chemistry teacher told us the formula of $\sum_{a=1}^{n}a^{2}$ is quite easy to remember. For e.g. when calculate $1^{2}+2^{2}+3^{2}... +5^{2}$ just multiply successively 5*6*(5+6)/6 that's it.

If $1^{2}+2^{2}+3^{2}... +7^{2}$ just multiply 7*8*(7+8)/6 successively.

5. ## Re: Arithmetic Progression

yeah. The formula is $\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}$