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

area = length x width

length = 2 x width

32 = 2 x width x width

32 = 2 x width^2

36 = width ^2  

width = 6

length = 12

area is 32

perimeter = 2 x 12 + 2 x 6 = 24 + 12 = 36

step-by-step explanation: