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How many terms of the geometric progression 1 + 4 + 16 + 64 + …….. must be added to get sum equal to 5461? - Mathematics

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Question

How many terms of the geometric progression 1 + 4 + 16 + 64 + …….. must be added to get sum equal to 5461?

Sum

Solution

Given G.P. : 1 + 4 + 16 + 64 + ...........

Here,

First term, a = 1

Common ratio, r = `4/1 = 4`  ...(∵ r > 1)

Let the number of terms to be added = n

Then, Sn = 5461

`=> (a(r^n - 1))/(r - 1) = 5461`

`=> (1(4^"th" - 1))/(4 - 1) = 5461`

`=>(4^"th" - 1)/3 = 5461`

`=>` 4th – 1 = 16383

`=>` 4th = 16384

`=>` n = 7

Hence, required number of terms = 7

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Simple Applications - Geometric Progression
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Chapter 11: Geometric Progression - Exercise 11 (D) [Page 161]

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Selina Mathematics [English] Class 10 ICSE
Chapter 11 Geometric Progression
Exercise 11 (D) | Q 2 | Page 161
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