Question

The 3rd and the 9th terms of a G.p are 54 and 39366 respectively find (a) common ratio, (b) 1st term, (c) 6th term and (d) sum of the first 10 terms

Answer 1

In GP n-th term is:

an = a • r ⁿ⁻¹

where

a = first term

r = common ratio

a3 = a • r ²

a3 = 54

a • r ² = 5r

a = 54 / r ²

a9 = a • r ⁸

a9 = 39366

a • r ⁸ = 39366

Replace a by 54 / r ² in this equation.

54 / r ² • r ⁸ = 39366

54 • r ⁶ = 39366

Divide both sides by 54.

r ⁶ = 729

r = six root ( 729 )

r = 3

Put this value in equation:

a • r ² = 54

a • 3 ² = 54

a • 9 = 54

a = 54 / 9

a = 6

a6 = a • r ⁵ = 6 • 3 ⁵ = 6 • 243

a6 = 1458

In GP sum of the first n terms is:

Sn = a ( 1 - r ⁿ ) / ( 1 - r )

In this case n = 10 , a = 6 , r = 3

S10= 6 • ( 1 - 3¹⁰ ) / ( 1 - 3 ) =

6 • ( 1 - 59049 ) / - 2 =

6 • ( - 59048 ) / - 2 =

- 354288 / - 2

S10 = 177144

Answer 2

3rd term = ar^2 = 54
9th term = ar^8 = 39366
divide them:
r^6 = 729 = 3^6
so r = 3

if ar^2 = 54 and r = 3
9a = 54
a = 6

use your definitions to find a, (did that)
r (did that), ar^5, and sum(10)

Answer 3

My typo:

It's not correct:

a • r ² = 5r

It should be written:

a • r ² = 54