The sum of an infinite G.P. is 3. The sum of the series formed by squaring its terms is also 3. The series isa)1/2 + 1/4 + 1/8 + 1/16 + ...b)3/2 + 3/4 + 3/8 + 3/16 + ...c)1/3 + 1/9 + 1/27 + 1/81 + ...d)1 - 1/3 + 1/32 - 1/33 + ...Correct answer is option 'B'. Can you explain this answer?

Question

Answer

Solution:

Given that the sum of an infinite G.P. is 3.

Let the first term be 'a' and the common ratio be 'r'.

The sum of this infinite GP is given by:

S = a/(1-r)

Given that the sum of the series formed by squaring its terms is also 3.

The sum of the series formed by squaring the terms is given by:

S' = a^2/(1-r^2)

Equating S and S', we get:

a/(1-r) = a^2/(1-r^2)

Simplifying the above equation, we get:

a = r/(1+r)

Substituting this value of 'a' in S, we get:

S = r/(1-r)

Simplifying the above equation using the value of 'a', we get:

S = (r/(1+r))^2 / (1-r^2)

Given that S = S' = 3.

Therefore,

(r/(1+r))^2 / (1-r^2) = 3

Simplifying the above equation, we get:

3r^4 + 6r^3 + 5r^2 + 2r - 3 = 0

This equation can be factorized as:

(r+1)(3r^3 + 3r^2 - 2r - 3) = 0

The first factor gives r = -1, which is not possible as it will make the infinite GP divergent.

Therefore, we need to solve the second factor:

3r^3 + 3r^2 - 2r - 3 = 0

This equation can be solved using numerical methods to get:

r = 1/2 or r = -1.292

The common ratio 'r' cannot be negative as it will make the infinite GP divergent.

Therefore, the common ratio 'r' is 1/2.

Hence, the infinite GP is:

3/2, 3/4, 3/8, 3/16, ...

Therefore, the correct option is (B).

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