The sum of an infinite geometric series is 4 and the sum of the cubes of the terms of the same GP is 192. The common ratio of the original geometric series is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(-\displaystyle\frac{1}{2}\)
An infinite GP with first term \(a\) and common ratio \(r\) (where \(-1\lt r\lt1\)) has sum \(\frac{a}{1-r}\). The cubes of its terms, \(a^3, a^3r^3, a^3r^6, \ldots\), form another GP with first term \(a^3\) and ratio \(r^3\).
The two sums:
\(\displaystyle\frac{a}{1-r}=4\), so \(a=4(1-r)\) (1)
\(\displaystyle\frac{a^3}{1-r^3}=192\) (2)
Putting \(a\) from (1) into (2), with \(1-r^3=(1-r)(1+r+r^2)\):
\(\displaystyle\frac{64(1-r)^3}{(1-r)(1+r+r^2)}=192\)
\((1-r)^2=3(1+r+r^2)\)
\(1-2r+r^2=3+3r+3r^2\)
\(2r^2+5r+2=0\)
\((2r+1)(r+2)=0\), so \(r=-\frac{1}{2}\) or \(r=-2\)
An infinite GP has a sum only when \(-1\lt r\lt1\), so \(r=-2\) is rejected and \(r=-\frac{1}{2}\) (with \(a=6\)).