Show that in an infinite G.P. with common ratio r (|r| < 1),

Question:

Show that in an infinite G.P. with common ratio r (|r| < 1), each term bears a constant ratio to the sum of all terms that follow it.

Solution:

Let us take a G.P. with terms $a_{1}, a_{2}, a_{3}, a_{4}, \ldots \infty$ and common ratio $r(|r|<1)$.

Also, let us take the sum of all the terms following each term to be $\mathrm{S}_{1}, \mathrm{~S}_{2}, \mathrm{~S}_{3}, \mathrm{~S}_{4}, \ldots$

Now, $\mathrm{S}_{1}=\frac{a_{2}}{(1-r)}=\frac{a r}{(1-r)}$

$S_{2}=\frac{a_{3}}{(1-r)}=\frac{a r^{2}}{(1-r)}$

$S_{3}=\frac{a_{4}}{(1-r)}=\frac{a r^{3}}{(1-r)}$

$\Rightarrow \frac{a_{1}}{S_{1}}=\frac{a}{\frac{a r}{(1-r)}}=\frac{(1-r)}{r}$

$\frac{a_{2}}{S_{2}}=\frac{a r}{\frac{a r^{2}}{(1-r)}}=\frac{(1-r)}{r}$

$\frac{a_{3}}{S_{3}}=\frac{a r^{2}}{\frac{a r^{3}}{(1-r)}}=\frac{(1-r)}{r}$

It is clearly seen that the ratio of each term to the sum of all the terms following it is constant.

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now