Single correct option
The sum of an infinite geometric progression is 24, and the sum of the squares of its terms is 192. What is the third term of the progression?
- A. 3/2
- B. 4
- C. 6
- D. 3
Show answer and solution
Answer: (D) 3
Let the first term be a and ratio r (|r| < 1).
a/(1 − r) = 24 … (i)
a²/(1 − r²) = 192 … (ii)
Divide (ii) by (i): a/(1 + r) = 8.
So a = 24(1 − r) = 8(1 + r) ⇒ 24 − 24r = 8 + 8r ⇒ r = 1/2, a = 12.
Third term = a·r² = 12 × 1/4 = 3. (6 is the second term — a trap.)
Tip: Dividing the two sum formulas cancels (1 − r) because 1 − r² = (1 − r)(1 + r).
a/(1 − r) = 24 … (i)
a²/(1 − r²) = 192 … (ii)
Divide (ii) by (i): a/(1 + r) = 8.
So a = 24(1 − r) = 8(1 + r) ⇒ 24 − 24r = 8 + 8r ⇒ r = 1/2, a = 12.
Third term = a·r² = 12 × 1/4 = 3. (6 is the second term — a trap.)
Tip: Dividing the two sum formulas cancels (1 − r) because 1 − r² = (1 − r)(1 + r).
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