JEE MAIN - Mathematics (2021 - 26th August Morning Shift - No. 11)
If the sum of an infinite GP a, ar, ar2, ar3, ....... is 15 and the sum of the squares of its each term is 150, then the sum of ar2, ar4, ar6, ....... is :
$${5 \over 2}$$
$${1 \over 2}$$
$${25 \over 2}$$
$${9 \over 2}$$
Explanation
Sum of infinite terms :
$${a \over {1 - r}} = 15$$ ..... (i)
Series formed by square of terms :
a2, a2r2, a2r4, a2r6 .......
Sum = $${{{a^2}} \over {1 - {r^2}}} = 150$$
$$ \Rightarrow {a \over {1 - r}}.{a \over {1 + r}} = 150 \Rightarrow 15.{a \over {1 + r}} = 150$$
$$ \Rightarrow {a \over {1 + r}} = 10$$ ...... (ii)
by (i) and (ii), a = 12; r = $${1 \over 5}$$
Now, series : ar2, ar4, ar6
Sum = $${{a{r^2}} \over {1 - {r^2}}} = {{12.\left( {{1 \over {25}}} \right)} \over {1 - {1 \over {25}}}} = {1 \over 2}$$
$${a \over {1 - r}} = 15$$ ..... (i)
Series formed by square of terms :
a2, a2r2, a2r4, a2r6 .......
Sum = $${{{a^2}} \over {1 - {r^2}}} = 150$$
$$ \Rightarrow {a \over {1 - r}}.{a \over {1 + r}} = 150 \Rightarrow 15.{a \over {1 + r}} = 150$$
$$ \Rightarrow {a \over {1 + r}} = 10$$ ...... (ii)
by (i) and (ii), a = 12; r = $${1 \over 5}$$
Now, series : ar2, ar4, ar6
Sum = $${{a{r^2}} \over {1 - {r^2}}} = {{12.\left( {{1 \over {25}}} \right)} \over {1 - {1 \over {25}}}} = {1 \over 2}$$
Comments (0)
