Unlike before, we cannot simply take the mean of each sequence and make them equal. The geometric means of the sequences would be equal, but we aren’t given the product of each sequence so that doesn’t help. As before we don’t know how many terms in the sequence, but we do know it must be of the form 6m+1.

The formula for the sum of a (finite) geometric sequence is:

S = a*(r^n-1)/(r-1) (r not equal to 1)

We are told our initial term is 1, so the a can be omitted from the formula.

For the main sequence the common ratio is simply r, and the number of terms n is 6m+1:

127 = (r^(6m+1)-1)/(r-1)

127(r-1) = r^6m*r – 1; let’s call r^6m = k

127r-126 = kr

k = 127-126/r

For our second sequence there are 3m+1 terms and the common ratio is r^2:

85 = (r^2(3m+1)-1)/(r^2-1)

85(r^2-1) = r^(6m+2)-1

85r^2-84 = kr^2

Let’s plug in the expression for k we found above:

85r^2-84 = (127-126/r)r^2

85r^2-84 = 127r^2-126r

42r^2 – 126r + 84 = 0

Solve the quadratic to find r = 1 or 2.

The formula for the sum of a geometric sequence specifies that r should not be exactly 1, but we’ll try it anyway if only to eliminate it as a possibility. The full sequence would just be 127 instances of the number 1. The second sequence would have 64 instances of 1 and therefore total 64, but this is not the case, so r=1 is not a valid solution.

If r = 2, the sequence goes 1,2,4,8 etc. The first 7 terms total 127. But if we want to be strictly systematic we know that k=127-126/2, so k=64. r^6m = k, so 2^6m = 64. m is therefore equal to 1. n, the number of terms in the sequence, is 6m+1, and therefore equal to 7.

Now to check the second sequence, 1+4+16+64=85, which is correct.

Finally to answer the question of the sum of the third sequence:

1+8+64 = 73.