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
A
\(\frac{860}{3}\)
B
\(\frac{680}{3}\)
C
\(\frac{608}{3}\)
D
\(\frac{680}{27}\)
Verified Correct Answer: B.
Step-by-Step Mathematical Solution
\(T_1\) = 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, \(S_4 = \frac{153(1 - (\frac{1}{3})^4)}{1 - \frac{1}{3}}\) = \(\frac{680}{3}\)
Source: Myschool Past Questions Archive (Record #32707)