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There are 6 numbers in geometric progression, sum of first three terms is 60 and last three terms is 128 find the first term and common ratio

i think ques is incomplete
i think, in place of 128, there should be 126.
please recheck the question


Let the number be , ar^-2,ar^-1,a,ar,ar^2 and ar^3
we got, ar^-2+ar^-1+a = 60
so , r^2+r+1 = 60r^2/a ----(i)

ar+ar^2+ar^3 = 128
ar(r^2+r+1) = 128 ------(2)
substituting (1)
gives common difference r= 2√2/∛15
and calculate the later, as I thnk the question is provided with wrong data
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