Single correct option
The sum of an infinite geometric progression is 16, and the sum of the squares of its terms is 256/3. What is the first term of the progression?
- A. 12
- B. 4
- C. 6
- D. 8
Show answer and solution
Answer: (D) 8
Let the first term be a and common ratio r. Then a/(1 − r) = 16 and a²/(1 − r²) = 256/3.
Divide the second by the square of the first: [a²/(1 − r²)] ÷ [a²/(1 − r)²] = (1 − r)/(1 + r) = (256/3)/256 = 1/3.
So 3 − 3r = 1 + r → r = 1/2, and a = 16 × (1 − 1/2) = 8.
Check: 64/(1 − 1/4) = 256/3 ✓.
Divide the second by the square of the first: [a²/(1 − r²)] ÷ [a²/(1 − r)²] = (1 − r)/(1 + r) = (256/3)/256 = 1/3.
So 3 − 3r = 1 + r → r = 1/2, and a = 16 × (1 − 1/2) = 8.
Check: 64/(1 − 1/4) = 256/3 ✓.
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