The sum of a given infinite geometric progression with a negative common ratio is 80 and the sum of its first two terms is 35. Then the value of \(n\) for which the sum of its first \(n\) terms is…
Question and options
The sum of a given infinite geometric progression with a negative common ratio is 80 and the sum of its first two terms is 35. Then the value of for which the sum of its first terms is closest to 100, is
- 4
- 5
- 7
- 6
Answer and explanation
Correct answer: B - 5
a/(1-r)=80 and a(1+r)=35 give 1-r^2=7/16, r=-3/4 (negative), a=140. S_n=80(1-(-3/4)^n): S4=54.69, S5=98.98, S6=65.76, S7=90.68. S5 is closest to 100, so n=5.
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