Question

Find the sum of first n terms of the series $3+7+13+21+31+....$

Or

If a b and c are in GP and x,y are the arithmetic means of a,b and b,c respectively. prove that $\frac{a}{x}+\frac{c}{y}=2and\frac{1}{x}+\frac{1}{y}=\frac{2}{b}$.

Answer 1

We have,the given series 3+7+13+21+31+...

The difference between the two successive terms are

$7-3=4,13-=6,21-13=8$

$31-21=10,$are in AP

Let $a_n$ be the nth terms and $S_n$ denotes the sum of n terms.

$Then,S_n=3+7+13+21+31+...+a_{n-1}+a_n..\left(i\right)$

$Also,S_n=3+7+13+21+...+a_{n-2}+a_{n-1}+a_n..\left(ii\right)$

On subtracting Eq.(ii)from Eq,(i),we get

$0=3+4+6+8+10+...+(a_n-a_{n-1})-a_n$

$\Rightarrow a_n=3+4+6+8+10+...+(a_n-a_{n-1})$

$\Rightarrow a_n=3+\frac{n-1}{2}[\because S_n=\frac{n}{2}[2a+(n-1)d\left]\right]$

$\Rightarrow a_n=3+\frac{n-1}{2}(8+2n-4)$

$\Rightarrow a_n=n^2+n+1$

$\therefore S_n=\sum a_n=\sum (n^2+n+1)$

$=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}+n$

$=n[\frac{(n+1)(2n+1)}{6}+\frac{(n+1)}{2}+1]$

$=n\left[\frac{(2n+3n+1)+3n+3+6}{6}\right]$

$=\frac{n}{6}[2n^2+6n+10]=\frac{n}{3}(n^2+3n+5)$

it is given that a,b and c are in GP.

$\therefore b^2=ac$

Also given x is the AM of a and b.

$\Rightarrow x=\frac{a+b}{2}$

and y is the AM of b and c.

$\Rightarrow y=\frac{b+c}{2}$

Now ,$\frac{a}{x}+\frac{c}{y}=\frac{a}{\frac{a+b}{2}}+\frac{c}{\frac{b+c}{2}}$

$=\frac{2a}{a+b}+\frac{2c}{b+c}$

$=\frac{2a(b+c)+2c(a+b)}{(a+b)(b+c)}$

$=\frac{2(ab+ac+ac+bc)}{ab+ac+b^2+bc}$

$=\frac{2(ab+2ac+b)}{ab+ac+a+bc}[\therefore b^2=ac]$

$=\frac{2(ab+2ac+bc)}{ab+2ac+bc}=2$

and $=\frac{1}{x}+=\frac{1}{y}=\frac{1}{=\frac{a+b}{2}}+\frac{1}{=\frac{b+}{2}}$

$=\frac{2}{a+b}+\frac{2}{b+c}=\frac{2(b+c+a+b)}{(a+b)(b+c)}$

$=\frac{2(a+c+2b)}{ab+b^2+ac+bc}$

$=\frac{2(a+c+2b)}{ab+b^2+b^2+bc}[\therefore ac=b^2]$

$=\frac{2(a+c+2b)}{b(a+2b+c)}=\frac{2}{b}$

$\therefore =\frac{1}{x}+\frac{1}{y}=\frac{2}{b}$ Hence proved.