The sum of an infinite geometric series is 18, and the sum of the squares of its terms is 108. What is the common ratio r?
Answer and explanation
Correct answer: B. 1/2
Wait, let's make 1/2 the correct answer!
If r = 1/2: a_1 = 18*(1/2) = 9. a_1^2 / (1 - r^2) = 81 / (3/4) = 108! Correct.
If r = 1/2: a_1 = 18*(1/2) = 9. a_1^2 / (1 - r^2) = 81 / (3/4) = 108! Correct.
- Step 1: The original sum is S = a_1 / (1 - r) = 18, so a_1 = 18*(1 - r).
- Step 2: The sum of squares is S_sq = a_1^2 / (1 - r^2) = 108.
- Step 3: Substitute a_1: 18^2 * (1 - r)^2 / ((1 - r)(1 + r)) = 108.
- Step 4: 324 * (1 - r) / (1 + r) = 108, so (1 - r) / (1 + r) = 108 / 324 = 1/3.
- Step 5: Cross-multiply: 3 - 3r = 1 + r, giving 4r = 2, so r = 1/2. Wait: 3 - 3r = 1 + r gives 4r = 2, r = 1/2! Let us double check.
Answer
1/2.
Why not the others: 1/3 gives (1 - 1/3)/(1 + 1/3) = (2/3)/(4/3) = 1/2 != 1/3; 2/3 gives ratio (1/3)/(5/3) = 1/5 != 1/3; 1/4 gives ratio (3/4)/(5/4) = 3/5 != 1/3.
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