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Question
Find the sum of the following geometric series:
`3/5 + 4/5^2 + 3/5^3 + 4/5^4 + ....` to 2n terms;
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Solution
Common Ratio = r = `(3/5)/(4/5^2) = 3/5 × 25/4 = 15/4`
∴ Sum of GP for n terms = `[a(r^n - 1)]/(r - 1)` ...(1)
⇒ a = `3/5, r = 15/4`, n = 2n
∴ Substituting the above values in (1), we get,
⇒ `[a(r^n - 1)]/(r - 1)`
⇒ `{3/5[(15/4)^"2n" - 1]}/(15/4 - 1)`
⇒ `{3/5[(15/4)^"2n" - 1]}/((15 - 1)/4)`
⇒ `{3/5[(15/4)^"2n" - 1]}/((11)/4)`
⇒ `{3[(15/4)^"2n" - 1]× 4}/(5 × 11)`
⇒ `{12[(15/4)^"2n" - 1]}/(55)`
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