In a geometric progression, the first term is 153 and the sixth term is \(\frac{17}{27}\)....

In a geometric progression, the first term is 153 and the sixth term is \(\frac{17}{27}\). The sum of the first four terms is

\(T_1\) = 153
\(T_6 = \frac{17}{27}\)
The equation for the nth term is ar\(^{n-1}\)
\(T_6: 153r^5\) = \(\frac{17}{27}\)

divide through by 153

r\(^5\) = \(\frac{17}{27}\) x \(\frac{1}{153}\)

take 5th root of both sides

r = \(\frac{1}{3}\)

The sum of n terms is \(S_n = \frac{a(1 - r^n)}{1 - r }\)
So, \(S_4 = \frac{153(1 - (\frac{1}{3})^4)}{1 - \frac{1}{3}}\) = \(\frac{680}{3}\)

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