Question

For a G.P. S4 = 81, S8 = 97, find r

Answer 1

Answer:

$\sf{The \ values \ of \ r \ are \ 1, \ -1, \ \dfrac{2}{3},} \\ \\ \sf{and \ -\dfrac{2}{3} \ respectively.}$

Given:

$\sf{In \ a \ G.P.,} \\ \\ \sf{\leadsto{S_{4}=81,}} \\ \\ \sf{\leadsto{S_{8}=97}}$

To find:

$\sf{The \ value \ of \ r.}$

Solution:

$\boxed{\sf{S_{n}=\dfrac{a(r^{n}-1)}{(r-1)}}} \\ \\ \sf{According \ to \ the \ first \ condition.} \\ \\ \sf{S_{4}=\dfrac{a(r^{4}-1)}{(r-1)}} \\ \\ \sf{\therefore{81=\dfrac{a(r^{4}-1)}{(r-1)}...(1)}} \\ \\ \sf{According \ to \ the \ second \ equation.} \\ \\ \sf{S_{8}=\dfrac{a(r^{8}-1)}{(r-1)}} \\ \\ \sf{\therefore{97=\dfrac{a(r^{8}-1)}{(r-1)}...(2)}} \\ \\ \sf{Divide \ equation (1) \ by \ equation (2), \ we \ get} \\ \\ \sf{\therefore{\dfrac{81}{97}=\dfrac{\frac{a(r^{4}-1)}{(r-1)}}{\frac{a(r^{8}-1)}{(r-1)}}}} \\ \\ \sf{\therefore{\dfrac{81}{97}=\dfrac{r^{4}-1}{r^{8}-1}}} \\ \\ \sf{\therefore{97r^{4}-97=81r^{8}-81}} \\ \\ \sf{\therefore{97r^{4}-81r^{8}-16=0}} \\ \\ \sf{Substitute \ r^{4}=n} \\ \\ \sf{81n^{2}-97n+16=0}$

$\sf{By \ quadratic \ formula, \ we \ get} \\ \\ \sf{n=\dfrac{97\pm\sqrt{97^{2}-4(81)(16)}}{162}} \\ \\ \sf{\therefore{n=\dfrac{97\pm\sqrt{4225}}{162}}} \\ \\ \sf{\therefore{n=\dfrac{97+65}{162} \ or \ \dfrac{97-65}{162} }} \\ \\ \sf{\therefore{n=\dfrac{162}{162} \ or \ \dfrac{32}{162}}} \\ \\ \sf{\therefore{n=1 \ or \ \dfrac{16}{81}}} \\ \\ \sf{Resubstituting \ n=r^{4}, we \ get} \\ \\ \sf{r^{4}=1 \ or \ \dfrac{16}{81}} \\ \\ \sf{\therefore{r=\pm1 \ or \ \pm\dfrac{2}{3}}} \\ \\ \purple{\tt{\therefore{The \ values \ of \ r \ are \ 1, \ -1,}}} \\ \\ \purple{\tt{\dfrac{2}{3}, \ -\dfrac{2}{3} \ respectively.}}$