Discussion
The sum of first three terms of a G.P. is 14 and the sum of next three terms is 112. Then 100th term of the G.P. is
Answer: C
The 100th term is \(2^{100}\).
Let the G.P. have first term \(a\) and ratio \(r\). Terms 4 to 6 are terms 1 to 3 multiplied by \(r^{3}\), so the ratio of the two sums gives \(r^{3}\) directly:
\[ \begin{aligned} r^{3} &= \frac{112}{14} = 8, \quad r = 2 \\ a(1 + 2 + 4) &= 14, \quad a = 2 \\ T_{100} &= a r^{99} = 2 \times 2^{99} = 2^{100} \end{aligned} \]
\(2^{99}\) is \(r^{99}\) without the factor \(a = 2\) (or the formula with \(r^{n}\) and \(a = 1\) mixed up); \(2^{101}\) uses \(r^{100}\). The options with \(-1\) and \(+1\) look like sum formulas, not a single term.
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