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Geometric progression — Statistics STD 11 Commerce — Question

Gujarat BoardEnglish MediumSTD 11 CommerceStatisticsGeometric progression2 Marks
Question
If in a G. P., $S_{n}=\frac{4}{3}\left(3^{n}-1\right)$, find $T_{3}$.
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Answer

We know $\quad \mathrm{T}_{n+1}=\mathrm{S}_{n+1}-\mathrm{S}_{n}$$\begin{aligned} \therefore \quad \mathrm{T}_{3} &=\mathrm{S}_{3}-\mathrm{S}_{2} \\ &=\frac{4}{3}\left(3^{3}-1\right)-\frac{4}{3}\left(3^{2}-1\right) \\ &=\frac{4}{3}[(27-1)-(9-1)] \\ &=\frac{4}{3}[26-8] \\ &=\frac{4}{3}(18) \\ \therefore \quad \mathrm{T}_{3} &=24 \end{aligned}$

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