If the sum of an infinite GP \(a, ar, ar^2, ar^3, \dots\) is 15 and the sum of the squares of its each term is 150, then the sum of \(ar^2, ar^4, ar^6, \dots\) is :
Show Hint
When terms of a GP are squared, the new GP has first term \(a^2\) and common ratio \(r^2\). Always check if the required sum starts from the first term or a later one.
Step 1: Understanding the Concept:
The sum of an infinite geometric progression (GP) with first term \(a\) and common ratio \(r\) (\(|r|<1\)) is \(S = \frac{a}{1-r}\). Step 2: Key Formula or Approach:
1. \(\frac{a}{1-r} = 15\)
2. \(\frac{a^2}{1-r^2} = 150\) Step 3: Detailed Explanation:
From (1), \(a = 15(1-r)\).
Substitute this into (2):
\[ \frac{[15(1-r)]^2}{(1-r)(1+r)} = 150 \implies \frac{225(1-r)}{1+r} = 150 \]
Divide by 75:
\[ \frac{3(1-r)}{1+r} = 2 \implies 3 - 3r = 2 + 2r \implies 5r = 1 \implies r = 1/5 = 0.2 \]
Then \(a = 15(1 - 0.2) = 15(0.8) = 12\).
The series required is \(ar^2, ar^4, ar^6, \dots\), which is an infinite GP with first term \(A = ar^2\) and common ratio \(R = r^2\).
\[ \text{Sum} = \frac{ar^2}{1-r^2} = \frac{12(0.2)^2}{1 - (0.2)^2} = \frac{12(0.04)}{1 - 0.04} = \frac{0.48}{0.96} = \frac{1}{2} \] Step 4: Final Answer:
The sum is \(1/2\).
