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Find the sum to indicated number of terms of the geometric progressions 7,21,37...n terms. - Mathematics

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Question

Find the sum to indicated number of terms of the geometric progressions `sqrt7, sqrt21,3sqrt7`...n terms.

Sum
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Solution

The geometric series `sqrt7, sqrt21, 3sqrt7,...`

First term, a = `sqrt7`

Common ratio, r = `sqrt21/sqrt7 = sqrt3`

Sum of n terms = `("a"(1 - "r"^"n"))/(1 - "r")` when r > 1

= `(sqrt7 [(sqrt3)^"n" - 1])/("r" -1)`

= `(sqrt7 [(sqrt3)^("n"/2) - 1])/(sqrt3 -1) xx (sqrt3 + 1)/(sqrt3 + 1)`

= `(sqrt7 (sqrt3 + 1)(3^("n"/2) - 1))/2`

= `(sqrt7(1 + sqrt3))/2[(3)^"n"/2 - 1]`

= `(sqrt7(1 + sqrt3))/2[(3)^(n/2) - 1]`

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Chapter 9: Sequences and Series - Exercise 9.3 [Page 192]

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NCERT Mathematics [English] Class 11
Chapter 9 Sequences and Series
Exercise 9.3 | Q 8 | Page 192

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