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

step-by-step explanation:

answer: all of the odd numbers were opened

the first person was the one that was opening all of the lockers. since one is odd, this causes all of the odd number lockers to be opened in the end.