Question

If the sum of an infinite GP \( a, ar, ar^2, ar^3, \dots \) is 15 and the sum of the squares of each term is 150, then the sum of the series \( ar^2, a r^4, ar^6, \dots \) is:

• A. \( \frac{5}{2} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{25}{2} \)
• D. \( \frac{9}{2} \)

Hint:

For infinite geometric progressions, remember: - The sum formula \( S = \frac{a}{1 - r} \) is valid for \( |r|<1 \). - The sum of squares follows \( S' = \frac{a^2}{1 - r^2} \). - Transforming a geometric progression into another requires adjusting the first term and common ratio accordingly.

Answer 1

The Correct Option is B

Step 1: Using the sum formula for an infinite geometric series: \[ S = \frac{a}{1 - r} \] Given that the sum of the infinite GP is 15, we get: \[ \frac{a}{1 - r} = 15 \quad \Rightarrow \quad a = 15(1 - r) \quad \cdots (1) \] 
Step 2: Using the sum of squares formula for an infinite GP: \[ S' = \frac{a^2}{1 - r^2} \] Given that the sum of the squares is 150, we get: \[ \frac{a^2}{1 - r^2} = 150 \quad \cdots (2) \] 
Step 3: Substituting \( a = 15(1 - r) \) into equation (2): \[ \frac{\left( 15(1 - r) \right)^2}{1 - r^2} = 150 \] \[ \frac{225 (1 - r)^2}{1 - r^2} = 150 \] Dividing both sides by 75: \[ \frac{3(1 - r)^2}{1 - r^2} = 2 \] Cross multiplying: \[ 3(1 - r)^2 = 2(1 - r^2) \] Expanding: \[ 3(1 - 2r + r^2) = 2 - 2r^2 \] \[ 3 - 6r + 3r^2 = 2 - 2r^2 \] \[ 5r^2 - 6r + 1 = 0 \] Solving for \( r \) using the quadratic formula: \[ r = \frac{6 \pm \sqrt{(-6)^2 - 4(5)(1)}}{2(5)} \] \[ r = \frac{6 \pm \sqrt{36 - 20}}{10} \] \[ r = \frac{6 \pm \sqrt{16}}{10} \] \[ r = \frac{6 \pm 4}{10} \] Possible values: \[ r = \frac{10}{10} = 1, \quad r = \frac{2}{10} = \frac{1}{5} \] Since \( |r|<1 \) for convergence, we take \( r = \frac{1}{5} \). 
Step 4: Finding \( a \) using equation (1): \[ a = 15(1 - \frac{1}{5}) = 15 \times \frac{4}{5} = 12. \] 
Step 5: Finding the sum of the new GP \( ar^2, ar^4, ar^6, \dots \), which forms another infinite GP with first term \( ar^2 \) and common ratio \( r^2 \): \[ S'' = \frac{ar^2}{1 - r^2} \] Substituting values: \[ S'' = \frac{12 \times \left(\frac{1}{5}\right)^2}{1 - \left(\frac{1}{5}\right)^2} \] \[ S'' = \frac{12 \times \frac{1}{25}}{1 - \frac{1}{25}} \] \[ S'' = \frac{\frac{12}{25}}{\frac{24}{25}} = \frac{12}{24} = \frac{1}{2} \]