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Finding the sum of the following series leads to two different answers with two different methods.

The series is listed as: 1+2/3 + 4/9 + 6/27 + …. (infinity)

Method 1: Directly apply the formula--> Sum=a/(1-r), with a=1 and r=2/3. Here, we get the answer as sum=3.

Method 2: First, simplify the series

                     S= 1+2/3 + 4/9 + 6/27 + …. (infinity)     …….(1)

                    3S = 3+2+ 4/3 + 6/9 + …...(infinity) .   ……..(2)

                     Subtracting (1) from (2) and solving further, we will get S=2.5

Kindly tell which method is correct.
ago in Numerical Ability by (5 points) | 23 views
+1
is it 6/27 or 8/27 ?

if it is 8/27 then those two methods should give the same answer.

Former one is very easy to apply.
0
The actual series is 1+2/3 + 4/9 + 6/27 + 8/81 …. (infinity)

Thanks a lot, I got to know where I was wrong
0
is it in GP ?

if so, what is r ?
0
The series is not in GP, I mistakenly considered 6/27 as 8/27 and thought it to be a GP series.

r is the common difference of the progression.
0
okay. So 2.5 is the answer.
0

how did you get 2.5 in 2nd method ?

+1
subtracted eqn 1 from eqn 2, obviously we will get 2.5, rt?

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