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Question

it is an easy question!

a geometric progression has first term k and last term h and the sum of all these term is S. find the common ratio in terms of k, h and S.

u almost got the answer but ........ it is wrong!!!

Answer 1

Hi,

The formula for the sum of geometric series is

Geometric sum form first to last term is = a{1-r^(n+1)}/(1-r)
=> {a-ar^(n+1)}/(1-r)

here a is the first term and hence is h
the last term is ar^(n+1) hence is k

Thus the required answer is Sum = (h-k)/(r-1)

Given the sum as S , ie., S = (h-k)/(r-1)

=> r-1 = (h-k)/S
=> r = {(h-k)/S} +1 = (h-k+S)/S

Peace out.

Answer 2

If the difference is r, and there are an n number of terms such that S is the sum,
S = k(1- r^n) / 1-r

Now kr^(n-1) = h
so, S =[ k - (h/k).r] / 1-r
S -Sr = k - (h/k).r
=> r[(h/k) - S] = k -S
Then,
r = (k-S)/[(h/k) -S] = k(k-S) / (h-Sk)