If the 2nd and 5th terms of a G.P are 6 and 48 respectively, find the sum of the first four for term
T\(_{2}\) = ar = 6
T\(_{5}\) = ar\(^{4}\) = 48
\(\frac{T_5}{T_2}\) = \(\frac{ar^{4}}{ar}\) = \(\frac{48}{6}\)
= r\(^{3}\) = 8
⇒ r = 2
S\(_{n}\) = \(\frac{a((r^n) - 1)}{r - 1}\)
S\(_{4}\) = \(\frac{a((r^4) - 1)}{r - 1}\)
S\(_{4}\) = \(\frac{3((2^4) - 1)}{2 - 1}\)
= 3(16 -1)
= 45
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