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Infinite Geometric Series

Easymathematics

Consider the infinite geometric series 9 - 6 + 4 - 8/3 + ... that converges; what is its exact sum to \infty?

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About This Question

Subject
mathematics
Chapter
sequence and series
Topic
infinite geometric series
Difficulty
Easy
Year
2025
Tags
advanced-calculus-drillinfinite-geometric-seriescommon-ratioconvergencealternating-series

Solution

Correct Answer:

An infinite geometric series converges precisely when the absolute value of its common ratio is strictly less than one, and its sum is then a/(1 - r), a cornerstone result tested at JEE Advanced. Here the first term is a = 9, and the ratio is r = -6/9 = -2/3, whose magnitude 2/3 is below one, guaranteeing convergence. The sum equals 9/(1 - (-2/3)) = 9/(5/3) = 27/5. Option 9/5 mistakenly uses r = +2/3 in the wrong sign. Option 5 misreads the first term as something other than 9. Option 27 ignores the denominator entirely and just scales the first term a. Hence the sum is 27/5. It helps to confirm the ratio by dividing any term by its predecessor: -6/9, 4/-6, and (-8/3)/4 all reduce to -2/3, so the geometric structure is genuine. Plausibility check: the partial sums 9, 3, 7, and 13/3 oscillate and progressively tighten around 5.4, and 27/5 = 5.4 lies between consecutive partial sums as expected for an alternating convergent series, confirming the answer.

This easy difficulty mathematics question is from the chapter sequence and series, covering the topic of infinite geometric series. It appeared in the 2025 exam.

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