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The sum to 5 terms of the GP 1 + 2/3 + 4/9 + … is
A243/211
B3 · (211/243)
C32/243
D211/81
Answer & Solution
Correct answer: D. 211/81
a = 1, r = 2/3. S_5 = (1 − (2/3)⁵)/(1 − 2/3) = 3·(211/243) = 211/81.
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