In a G. P., the third term exceeds the first term by 16.

If the sum of the third term and the fourth term is 72,

find the common ratios.

Step-by-step explanation:

In a G. P, the nth term is given as

Un=ar^ (n-1)

Where

a is first term

n is nth term

And r is common ratio

So in the question given above,

The third term exceed the first term by 16

i. e U3=U1+16

Where U1=a. First term

U3=a+16

Given also that, the sum if the third term and fourth term is 72.

Then U3+U4=72.

We are told to find common ratio (r)

U3=ar^3-1

U3=ar^2

Also, U4=ar^3

U3+U4=72

ar^2+ar^3=72

ar^2 (1+r) = 72. equation 1

Also for

U3=a+16

ar^2=a+16

ar^2-a=16

a (r^2-1) = 16. From (x^2-y^2) = (x+y) (x-y)

Then,

a (r-1) (r+1) = 16. Equation 2

Divide equation 2 by equation 1

a (r-1) (r+1) / ar^2 (1+r) = 16/72

Then a cancel a and (1+r) cancel (1+r)

So,

(r-1) / r^2=2/9

Cross multiply

9 (r-1) = 2r^2

2r^2-9r+9=0

Solving the quadratic equation

2r^2-6r-3r+9=0

2r (r-3) - 3 (r-3) = 0

(r-3) (2r-3) = 0

r-3=0. Or. 2r-3 = 0

Then r=3 or r=3/2