if sum of four terms of GP is 60 and am between first and last term is 18. then find a and r.

See this solution and answer me if you are satisfied.

Let the numbers be a , ar , ar^2 , ar^3 .

Now, sum=60.

=> a + ar + ar^2 +ar^3 =60.    ...........(i).

Also, a + ar^3 = 18*2 = 36.      ...........(ii).

=> a ( r + r^3 ) = 36.

=> a = 36 / ( r + r^3).

Now putting this in eq.(i) , we get 24(r^3) - 36(r^2) - 36r +24 = 0.

=> r = 1/2 , -1 , 2.

r=1 doesn't satisfy the condition.

So r=2 or r=1/2

When r=2 , a=4 , and when r=1/2 , a=32. Hence, solved.

Dear Student,
Please find below the solution to your problem.

Let the numbers be a , ar , ar^2 , ar^3 .
Now, sum=60.
=> a + ar + ar^2 +ar^3 =60. ...........(i).
Also, a + ar^3 = 18*2 = 36. ...........(ii).
=> a ( r + r^3 ) = 36.
=> a = 36 / ( r + r^3).
Now putting this in eq.(i) , we get 24(r^3) - 36(r^2) - 36r +24 = 0.
=> r = 1/2 , -1 , 2.
r=1 doesn't satisfy the condition.
So r=2 or r=1/2
When r=2 , a=4 , and when r=1/2 , a=32.

Thanks and Regards