Discussion
The sum of four consecutive terms in a geometric progression is 960. If the fourth term is 8 times as large as the first term, then the smallest number in the geometric progression is
Answer: B
The smallest number is 64.
Let the four terms be \(a, ar, ar^{2}, ar^{3}\). The fourth term is 8 times the first, so \(r^{3} = 8\) and \(r = 2\).
\[ \begin{aligned} a(1 + 2 + 4 + 8) &= 960 \\ 15a &= 960 \\ a &= 64 \end{aligned} \]
The terms are 64, 128, 256 and 512, which add to 960. Since the ratio is greater than 1, the first term is the smallest. The option 128 is the second term, and 240 is \(960 \div 4\), the average rather than a term.
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