Question

The 6th term of a G.P. is 1215. Given that the
common ratio is 3, find the 9th term.​

Answer 1

Answer:

Let A be the first term and r be the common ration of the sequence.

Then Ar^2 = 360 and Ar^5 = 1215.

So r^3 = 1215 / 360 = 3.375

r = cube root (3.375) = 1.5

So A = 360/1.5^2 = 360/2.25 = 160.

So the sum of the first four terms = 160 * (1.5^4 -1)/(1.5–1)

= 320 * (1.5^4 -1) = 1,300.