Question

Find the maximum value of n such that the sum of the first n terms of a G.P. 1, 3, 32, 3 … does not exceed 365.

Answer 1

Here, a = 1; r = 31 = 3; The sum of the first n terms does not exceed 365, i.e.,

S_n ≤ 365.

Now, S_n = r-i

Putting a = 1; r = 3.

S_n = \(\frac{1[3_n−1]}{3−1}\)

S_n = \(\frac{3^n−1}{2}\)

Now, S_n ≤ 365

∴ \(\frac{3^n−1}{2}\) ≤ 365

∴3ⁿ – 1≤ 365 × 2

∴ 3ⁿ ≤ 730 + 1

∴ 3ⁿ ≤ 731

We now tabulate values of 3’ for different positive values of n as shown in the following table and we shall take the maximum value of n for which 3ⁿ ≤ 731.

n	5	6	7
3n	243	729	2187

For n = 7, 3ⁿ = 2187 which is more than 731.

∴ n = 6 is the maximum value of n, for which 3ⁿ ≤ 731.