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Sequences and Series (p-1) — Maths (commerce) STD 11 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 11 Commerce / ArtsMaths (commerce)Sequences and Series (p-1)3 Marks
Question
For a sequence, $S_n=4\left(7^n-1\right)$, verify whether the sequence is a G.P.
✓

Answer

$
\begin{aligned}
& \mathrm{S}_{\mathrm{n}}=4\left(7^{\mathrm{n}}-1\right) \\
& \therefore \quad \mathrm{S}_{\mathrm{n}-1}=4\left(7^{\mathrm{n}-1}-1\right) \\
& \text { But, } \mathrm{t}_{\mathrm{n}}=\mathrm{S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}-1} \\
& =4\left(7^n-1\right)-4\left(7^{n-1}-1\right) \\
& =4\left(7^{\mathrm{n}}-1-7^{\mathrm{n}-1}+1\right) \\
& =4\left(7^{\mathrm{n}}-7^{\mathrm{n}-1}\right) \\
& =4\left(7^{\mathrm{n}-1+1}-7^{\mathrm{n}-1}\right) \\
& =4.7^{\mathrm{n}-1}(7-1) \\
& \therefore \quad \mathrm{t}_{\mathrm{n}}=24.7^{\mathrm{n}-1} \\
& \therefore \quad \mathrm{t}_{\mathrm{n}-1}=24.7^{(\mathrm{n}-1)-1}=24.7^{\mathrm{n}-2} \\
&
\end{aligned}
$
The sequence is a G.P., if $\frac{t_n}{t_{n-1}}=$ constant for all $\mathrm{n} \in \mathrm{N}$.
$
\begin{aligned}
\therefore \quad \frac{\mathrm{t}_{\mathrm{n}}}{\mathrm{t}_{\mathrm{n}-1}} & =\frac{24.7^{\mathrm{n}-1}}{24.7^{\mathrm{n}-2}}=\frac{7^{\mathrm{n}-1}}{7^{\mathrm{n}-1} \cdot 7^{(-1)}} \\
& =7=\text { constant, for all } \mathrm{n} \in \mathrm{N}
\end{aligned}
$
$\therefore \quad$ the sequence is a G.P.

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