Question

If Sn denotes the sum to n terms of a GP show that (S10-S20)^2=S10(S30-S20)

Answer 1

Since, $S_n$ denotes the sum to 'n' terms of GP.

We have to prove that $(S_{10}-S_{20})^2 = S_{10}(S_{30} - S_{20})$

Sum to 'n' terms of GP with first term 'a' and common ratio 'r' is $\frac{a(1-r^n)}{1-r}$

Consider LHS

= $(S_{10}-S_{20})^2$

= $(\frac{a(1-r^{10})}{1-r} - \frac{a(1-r^{20})}{1-r})^2$

= $(\frac{a(1-r^{10}) -a(1-r^{20})}{1-r})^2$

= $(\frac{a-ar^{10} -a+ar^{20}}{1-r})^2$

= $(\frac{ar^{20}-ar^{10}}{1-r})^2$

= $a^2(r^{10}[\frac{r^{10}-1}{1-r}])^2$

= $a^2r^{20}(\frac{r^{10}-1}{1-r})^2$

= $a^2r^{20}(\frac{1-r^{10}}{1-r})^2$

Consider RHS

= $S_{10}(S_{30}-S_{20})$

= $\frac{a(1-r^{10})}{1-r} [\frac{a(1-r^{30})}{1-r} - \frac{a(1-r^{20})}{1-r}]$

= $\frac{a(1-r^{10})}{1-r} [ \frac{a-ar^{30}-a+ar^{20}} {1-r}]$

=$\frac{a(1-r^{10})}{1-r} [ \frac{ar^{20}-ar^{30}} {1-r}]$

= $\frac{a^2(1-r^{10})}{1-r} [r^{20} (\frac{1-r^{10}} {1-r})]$

= $a^2r^{20}(\frac{1-r^{10}}{1-r})^2$

So, LHS = RHS

Hence proved.

Answer 2

Step-by-step explanation:

Hope it helps you .......