# The 4th and 7th terms of a G.P.

Question:

The 4 th and 7 th terms of a G.P. are $\frac{1}{27}$ and $\frac{1}{729}$ respectively. Find the sum of $n$ terms of the G.P.

Solution:

Let a be the first term and r be the common ratio of the G.P.

$\therefore a_{4}=\frac{1}{27}$

$\Rightarrow a r^{4-1}=\frac{1}{27}$

$\Rightarrow a r^{3}=\frac{1}{27}$

$\Rightarrow\left(a r^{3}\right)^{2}=\frac{1}{27^{2}}$

$\Rightarrow a^{2} r^{6}=\frac{1}{729}$

$\Rightarrow a r^{6}=\frac{1}{729 a}$    ...(i)

Similarly, $\mathrm{a}_{7}=\frac{1}{729}$

$\Rightarrow a r^{7-1}=\frac{1}{729}$

$\Rightarrow a r^{6}=\frac{1}{729}$

$\Rightarrow a r^{6}=\frac{1}{729 \mathrm{a}}$     [From (i)]

Putting this in $a_{4}=\frac{1}{27}$

$\Rightarrow a r^{3}=\frac{1}{3^{3}}$

$\Rightarrow r^{3}=\frac{1}{3^{3}}$

$\therefore r=\frac{1}{3}$

Now, sum of $n$ terms of the G.P., $S_{n}=a\left(\frac{r^{n}-1}{r-1}\right)$

$\Rightarrow S_{n}=1\left(\frac{1-\left(\frac{1}{3}\right)^{n}}{1-\frac{1}{3}}\right)$

$\Rightarrow S_{n}=\frac{3}{2}\left(1-\frac{1}{3^{n}}\right)$

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