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15, 30, 60, 120.... are in G.P. (Geometric Progression): (a) Find the nth term of this G.P. in terms of n. (b) How many terms of the above G.P. will give the sum 945? - Mathematics

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Question

15, 30, 60, 120.... are in G.P. (Geometric Progression):

  1. Find the nth term of this G.P. in terms of n.
  2. How many terms of the above G.P. will give the sum 945?
Sum
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Solution

a. Given, G.P. is 15, 30, 60, 120....

Here, a = 15

Common ratio (r) = `30/15` = 2

Then an = arn – 1

= 15(2)n – 1

b. Sum of n terms,

`S_n = (a(r^n - 1))/(r - 1)`   ...(∵ r > 1)

`\implies 945 = 15((2^n - 1))/(2 - 1)`

`\implies 945/15 = 2^n - 1`

`\implies` 63 = 2n – 1

`\implies` 63 + 1 = 2n

`\implies` 64 = 2n

`\implies` 2n = 64

`\implies` 2n = 26

∴ n = 6

Hence, the number of terms needed is 6.

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Chapter 9: Arithmetic and geometric progression - Exercise 9E [Page 199]

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Nootan Mathematics [English] Class 10 ICSE
Chapter 9 Arithmetic and geometric progression
Exercise 9E | Q 13. | Page 199
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