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Find the first term and common ration of a geometric progression if the sum to n is given by

6 - 2/(3^(n-1))

I tried solving by making both terms of the eqn having the same denominator by multiplying (3^(n-1)) to the first term 6 and then by taking out the 2, i am able to make the formula to 2(3^n - 1)/ (3^n-1). But what about the denominator, i remember that there must be no n at the bottom. Only r - 1 where in this case, r is to be found..

Anyone out here can give me some guide? Thanks for any help given.