A geometric sequence has first term and common ratio . What is the smallest such that the sum of the first terms exceeds ?
Answer and explanation
Correct answer: D.
- Step 1: .
- Step 2: We need , so .
- Step 3: is too small and works, so (sum ).
Answer
.
Why not the others: : gives a sum of , which is still below . : gives a sum of , but terms already exceed . : gives a sum of , well below .
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