1. ## Arithmetic Progression

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

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

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

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

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

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

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

When n=1

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

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

In equation (i)

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

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

2. ## Re: Arithmetic Progression

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

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

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

I'm trying to get used to TEX

3. ## Re: Arithmetic Progression

Thanks, I got it.

$\displaystyle a_{n}=2cn-c$

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

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

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

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

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

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

So the correct answer is (c).

Am I correct?

4. ## Re: Arithmetic Progression

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

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

5. ## Re: Arithmetic Progression

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