Question

Find the minimum value of n such that the sum of the first n terms of a G.P. 1, 2, 22, 2, … is greater than or equal to 2000.

Answer 1

Here, a = 1; r = 21 = 2, The sum of first ii terms is greater than or equal to 2000, i.e..

S_n ≥ 2000.

Now, S_n = \(\frac{a[r^n−1]}{r−1}\)

Putting a = 1; r = 2,

S_n = \(\frac{1[2^n−1]}{r−1}\)= 2ⁿ – 1

S_n ≥ 2000

∴ 2ⁿ – 1 ≥ 2000

∴ 2ⁿ ≥ 2000 + 1

∴ 2ⁿ ≥ 2001

We now tabulate values of 2ⁿ for different positive values of n as shown in the following table and we shall take the minimum value of n for which 2ⁿ ≥ 2001.

n	8	9	10	11	12
2n	256	512	1024	2048	4096

From the table we can see that for n = 11, 2ⁿ = 2048 which is greater than 2001.

∴ For n = 11, 2n 2001.

Hence, n = 11 is minimum value of n.