If 7 and 189 are the first and fourth terms of geometric progression respectively, find the sum of the first three terms of the progression
\(T_{n} = ar^{n - 1}\) (nth term of a G.P)
\(T_{4} = ar^{3} = 189\)
\(7 \times r^{3} = 189 \implies r^{3} = 27\)
\(r = \sqrt[3]{27} = 3\)
\(S_{n} = \frac{a(r^{n} - 1)}{r - 1}\)
\(S_{3} = \frac{7(3^{3} - 1)}{3 - 1} \)
= \(\frac{7 \times 26}{2} = 91\)
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