Probably we need to find all the terms of this progression.
Recall that each next term of a geometric progression is obtained by multiplying the previous term by the quotient, denote it as `q.` Denote the first term as `b,` then the second is `b*q,` the third is `b*q*q = b*q^2,` and the k-th term is `b*q^(k-1).`
It is also well-known that the sum of `n` terms of a geometric progression with the first term `b` and `q!=1` is `S = b*(q^n-1)/(q-1).`
In our case `q=2,` `b` is unknown, `n = 7` and `S = 127.` This way we obtain a simple equation for `b:`
`127 = b*(2^7-1)/(2-1) = b*127,` because `2^7 = 128` (check by multiplying 2*2*2*2*2*2*2).
Obviously the only solution is `b=1,` and the entire array is 1, 2, 4, 8, 16, 32, 64.
We could solve this problem without the formula for sum, adding all `7` terms manually:
`b+2b+4b+8b+16b+32b+64b = 127,` the left side is `(1+2+4+8+16+32+64)b = 127b = 127,` and again `b=1.`