मराठी

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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प्रश्न

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.

बेरीज
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उत्तर

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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पाठ 8: Sequences and Series - Miscellaneous Exercise [पृष्ठ १४७]

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एनसीईआरटी Mathematics [English] Class 11
पाठ 8 Sequences and Series
Miscellaneous Exercise | Q 5. | पृष्ठ १४७

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