JEE MAIN - Mathematics (2002 - No. 52)
Sum of infinite number of terms of GP is 20 and sum of their square is 100. The common ratio of GP is
5
3/5
8/5
1/5
Explanation
Let $$a=$$ first team of $$G.P.$$ and $$r=$$ common ratio of $$G.P.;$$
Then $$G.P.$$ is $$a,$$ $$ar,$$ $$a{r^2}$$
Given $${S_\infty } = 20 \Rightarrow {a \over {1 - r}} = 20$$
$$ \Rightarrow a = 20\left( {1 - r} \right)....\left( i \right)$$
Also $${a^2} + {a^2}{r^2} + {a^2}{r^4} + ...$$ to $$\infty = 100$$
$$ \Rightarrow {{{a^2}} \over {1 - {r^2}}} = 100$$
$$ \Rightarrow {a^2} = 100\left( {1 - r} \right)\left( {1 + r} \right)....\left( {ii} \right)$$
From $$(i),$$ $${a^2} = 400{\left( {1 - r} \right)^2};$$
From $$(ii),$$ we get $$100\left( {1 - r} \right)\left( {1 + r} \right)$$
$$\,\,\,\,\,\,\,\,\,\, = 400{\left( {1 - r} \right)^2}$$
$$ \Rightarrow 1 + r = 4 - 4r$$
$$ \Rightarrow 5r = 3$$
$$ \Rightarrow r = 3/5.$$
Then $$G.P.$$ is $$a,$$ $$ar,$$ $$a{r^2}$$
Given $${S_\infty } = 20 \Rightarrow {a \over {1 - r}} = 20$$
$$ \Rightarrow a = 20\left( {1 - r} \right)....\left( i \right)$$
Also $${a^2} + {a^2}{r^2} + {a^2}{r^4} + ...$$ to $$\infty = 100$$
$$ \Rightarrow {{{a^2}} \over {1 - {r^2}}} = 100$$
$$ \Rightarrow {a^2} = 100\left( {1 - r} \right)\left( {1 + r} \right)....\left( {ii} \right)$$
From $$(i),$$ $${a^2} = 400{\left( {1 - r} \right)^2};$$
From $$(ii),$$ we get $$100\left( {1 - r} \right)\left( {1 + r} \right)$$
$$\,\,\,\,\,\,\,\,\,\, = 400{\left( {1 - r} \right)^2}$$
$$ \Rightarrow 1 + r = 4 - 4r$$
$$ \Rightarrow 5r = 3$$
$$ \Rightarrow r = 3/5.$$
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