Question
Sum of infinite number of terms of G.P. is 20 and sum of their square is 100. The common ratio of G.P. is
A.
5
B.
$$\frac{3}{5}$$
C.
$$\frac{8}{5}$$
D.
$$\frac{1}{5}$$
Answer :
$$\frac{3}{5}$$
Solution :
Let $$a$$ = first term of G.P. and $$r$$ = common ratio of G.P.;
Then G.P. is $$a,ar,a{r^2}$$
$$\eqalign{
& {\text{Given }}{S_\infty } = 20 \Rightarrow \frac{a}{{1 - r}} = 20 \cr
& \Rightarrow a = 20\left( {1 - r} \right)\,\,\,\,\,\,\,\,.....\left( {\text{i}} \right) \cr
& {\text{Also }}{a^2} + {a^2}{r^2} + {a^2}{r^4} + ..... + {\text{to}}\,\infty = 100 \cr
& \Rightarrow \,\,\frac{{{a^2}}}{{1 - {r^2}}} = 100 \cr
& \Rightarrow {a^2} = 100\left( {1 - r} \right)\left( {1 + r} \right)\,\,\,\,\,.....\left( {{\text{ii}}} \right) \cr
& {\text{From }}\left( {\text{i}} \right),{a^2} = 400{\left( {1 - r} \right)^2}; \cr
& {\text{From }}\left( {{\text{ii}}} \right),{\text{ we get 100}}\left( {1 - r} \right)\left( {1 + r} \right) = 400{\left( {1 - r} \right)^2} \cr
& \Rightarrow \,\,1 + r = 4 - 4r \cr
& \Rightarrow \,\,5r = 3 \cr
& \Rightarrow \,\,r = \frac{3}{5}. \cr} $$