Question

if S1, S2 & S3 represent the number of terms 2n terms, 3n of the first term geometry series prove that S1 (S3-S2)=(S2-S1)^2​

Answer 1

Step-by-step explanation:

Given :

$S_1,S_2,\ and \ S_3$ represent the sum of first n terms, 2n terms, and 3n terms of a G.P,.

To Prove :

$S_1(S_3 - S_2) = (S_2 - S_1)^2$

Solution :

Proof :

Let the first term be "a",

and the common ratio be "r" , then sum of first n terms of the series will be found by the formula,

⇒ $S_n=a(\frac{r^n - 1}{r-1})$

∴ $S_1 = a(\frac{r^n - 1}{r-1})$

∴ $S_2 = a(\frac{r^{2n} - 1}{r-1})$

∴ $S_3 = a(\frac{r^{3n} - 1}{r-1})$

Hence,

LHS = $S_1(S_3 - S_2)$

= $a(\frac{r^n - 1}{r-1}) [a(\frac{r^{3n} - 1}{r-1}) -a(\frac{r^{2n} - 1}{r-1})]$

=$a(\frac{r^n - 1}{r-1}) \frac{a}{r-1} [(r^{3n} -1 -(r^{2n} - 1)]$

= $(\frac{a}{r-1})(r^n - 1)(\frac{a}{r-1}) [(r^{3n}-r^{2n}]$

= $(\frac{a}{r-1})^2(r^n-1)r^{2n}(r^n - 1)$

= $(\frac{a}{r-1})^2(r^n-1)^2r^{2n}$

= $(\frac{a}{r-1})^2(r^n-1)^2(r^{n})^2$

= $[(\frac{a}{r-1})(r^n-1)(r^{n})]^2$

= $[(a(\frac{r^n-1}{r-1})(r^{n})]^2$

= $[a(\frac{r^n-1}{r-1}){[1 - (r^{n} + 1)]^2$

= $[a(\frac{r^n-1}{r-1}) - (a(\frac{r^n-1}{r-1}))(r^{n} + 1)]^2$

= $[a(\frac{r^n-1}{r-1}) - (a(\frac{(r^n-1)(r^{n} + 1)}{r-1}))]^2$

= $[a(\frac{r^n-1}{r-1}) - (a(\frac{(r^{2n}-1)}{r-1}))]^2$

= $[a(\frac{r^{2n}-1}{r-1}) - a(\frac{r^n-1}{r-1})]^2$

= $(S_2 - S_1)^2$ = RHS