Question

The fith term of a GP is greater than the 4th term by 13 and 1\2and the 4th term is greater than the 3rd term by 9 find the sum to the 5th term of the sequence

Answer 1

term5 - term4 = 13 1/2
ar^4 - ar^3 = 27/2
ar^3(r - 1) = 27/2 **

term4 - term3 = 9
ar^3 - ar^2 = 9
ar^2(r-1) = 9 ***

divide ** by ***
r = (27/2) / 9 = 27/18 = 3/2

back into ***
ar^2(r-1) = 9
a(9/4)(3/2 - 1) = 9
a(9/4)(1/2) = 9
a = 8

your terms are: 8, 12, 18, 27, 40.5,
check:
term5 - term4 = 40.5 - 27 = 13.5
term4 - term3 = 27-18 = 9
my terms are correct

sum(5) = a(r^5 - 1)/(r-1)
= 8( (3/2)^5 - 1)/(1/2)
= 16(243/32 - 1)
= 16(211/32) = 211/2 or 105.5

we could have just as well added up the terms stated above