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Question
A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.
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Solution
Let the first term of the geometric progression = a, common ratio = r and number of terms = 2n.
Sum of all terms = `("a"("r"^(2"n") - 1))/("r" - 1)`
Terms placed at odd places a, ar2, ar4,…. up to n terms
Their sum = a + ar2 + ar2 +…… up to n terms
= `("a"[("r"^2)^"n" - 1])/("r"^2 - 1) = ("a"("r"^(2"n") - 1 ))/("r"^2 - 1)`
Given:
Sum of 2n terms of a geometric series = 5 × [Sum of terms at odd places]
⇒ `("a"("r"^(2"n") - 1))/("r" - 1) = 5 xx ("a"[("r"^2)^"n" - 1 ])/("r"^2 - 1)`
or `("a"("r"^(2"n") - 1))/("r" - 1) = (5"a"("r"^(2"n") - 1)) /("r"^2 - 1)`
`1 = 5/("r" + 1)`
or r + 1 = 5
or r = 4
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