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Question
Find the sum to infinity of the following arithmetico - geometric sequence:
`3, 6/5, 9/25, 12/125, 15/625, ...`
Solution
S = `3 + 6/5 + 9/5^2 + 12/5^3 + 15/5^4 + ...` ...(i)
Multiplying (i) by `1/5`, we get
`1/5"S" = 3/5 + 6/5^2 + 9/5^3 + 12/5^4 + 15/5^5 + ...` ...(ii)
Equation (i) – (ii), we get
`4/5"S" = 3 + (3/5 + 3/5^2 + 3/5^3 + ...)`
The terms `3/5, 3/5^2, 3/5^3` are in G.P.
∴ a = `3/5`, r = `1/5`
Since, |r| = `|1/5| < 1`
∴ sum to infinity exists.
∴ `4/5"S" = 3 + (3/5)/(1 - 1/5)`
= `3 + 3/4`
∴ `4/5"S" = 15/4`
∴ S = `75/16`
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