If 4th term is 27, \(ar^3=27....1*r^3=27....r=3\)

Sum of first n terms of a GP = \(\frac{a(r^n-1)}{r-1}=\frac{1(3^7-1)}{3-1}=\frac{2187-1}{2}=\frac{2186}{2}=1093\)

A

You don't need to know what a "geometric progression" is on the GMAT, and you certainly don't need to know any geometric sum formulas - where is the question from?

In a geometric sequence, we find the next term by multiplying the previous term by a constant. If the first term is a, and our constant is r, then the second term is ar, the third is ar^2, the fourth is ar^3, and so on. So here if the first term is 1, and the fourth is 27, then r^3 = 27 and r = 3. So the first seven terms are

1, 3, 9, 27, 81, 243, 729

and just quickly estimating, only answer A makes any sense (A also is the only answer with the right units digit).
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