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Sequences and Series question

2002 · Shift 0 · Q113

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2002 · Shift 0 · Q113

Sequences and Series question

2002 · Shift 0 · Q113

JEE MainMathematicsSequences and SeriesMCQ+4 / −1

Sum of infinite number of terms of GP is 20 and sum of their square is 100. The common ratio of GP is

A
5
B
3/5
C
8/5
D
1/5

View written solutionFree

Correct answer: B

Let the first term of the GP be $a$ and common ratio be $r$ .

Since the sum of infinitely many terms exists, we must have $|r|<1.$
Given that the sum to infinity is $20$ : $\frac{a}{1-r}=20 \quad \Rightarrow \quad a=20(1-r).$
The terms of the GP are $a,\ ar,\ ar^2,\ ar^3,\dots$ Their squares form another GP: $a^2,\ a^2r^2,\ a^2r^4,\dots$ whose sum to infinity is given as $100$ .

So, $\frac{a^2}{1-r^2}=100.$
Substitute $a=20(1-r)$ into this equation: $\frac{[20(1-r)]^2}{1-r^2}=100.$

$\frac{400(1-r)^2}{(1-r)(1+r)}=100.$

Simplifying, $\frac{400(1-r)}{1+r}=100.$

Divide by $100$ : $\frac{4(1-r)}{1+r}=1.$
Solve for $r$ : $4(1-r)=1+r$ $4-4r=1+r$ $3=5r$ $r=\frac{3}{5}.$
Check the condition: $\left|\frac{3}{5}\right|<1,$ so this is valid.
Option check:

A: $5$ is invalid since $|r|<1$ is required.
B: $\frac{3}{5}$ is correct.
C: $\frac{8}{5}$ is invalid since $|r|<1$ is required.
D: $\frac{1}{5}$ does not satisfy the equations.

Therefore, the common ratio is $\boxed{\frac{3}{5}}.$

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