In a geometric progression with terms, the sum of the first and last term is 66 and the product of the second and second last term is 128. Given that the sum of all the terms of this geometric progression is 126, find the number of terms and possible values of common ratio.

Answer 1

So, suppose there are n terms. You know that
a + ar^(n-1) = 66
ar * ar^(n-2) = 128
a(r^n - 1)/(r-1) = 126
You can go through the math, but 128 is a product of powers of 2, so just looking at it, I'd guess the sequence is
2,4,8,16,32,64