Infinite Geometric Series

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.