Answer:
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
