Gujarat BoardEnglish MediumSTD 11 CommerceStatisticsGeometric progression2 Marks
Question
If for a G.P., $\mathbf{T}_{n}=2^{n+1}$, obtain $\mathrm{S}_{4}$.
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Answer
Here, $\mathrm{T}_{n}=2^{n+1}$$\therefore \quad \mathrm{T}_{1}=2^{1+1}=4, \mathrm{~T}_{2}=2^{2+1}=8, \mathrm{~T}_{3}=2^{3+1}=16, \ldots$Here, the first term $a=4$ and the common ratio $r=\frac{8}{4}=2 .$ Sum of the first four terms is required i.e. $n=4$.Putting the values of $a, r$ and $n$ in $\mathrm{S}_{n}=\frac{a\left(r^{n}-1\right)}{(r-1)}$, we get$\begin{aligned}\mathrm{S}_{4} &=\frac{4\left(2^{4}-1\right)}{(2-1)} \\&=\frac{4(16-1)}{1} \\&=4 \times 15 \\&=60\end{aligned}$Thus, the sum of the first four terms of the G. P. is 60 .
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