# Sequences and series

• April 12th 2006, 06:26 AM
FuNkY14
Sequences and series
The length of the radii of circles form an infinite geameric sequence. The length of the radius of the first circle is 6cm. The length of the radius of each of the circles is 4/5 of the length of the radius of the previous circle. Show that the total area of all the circles formed in this way is 100pi cm2.

FuNk
• April 12th 2006, 07:01 AM
CaptainBlack
Quote:

Originally Posted by FuNkY14
The length of the radii of circles form an infinite geameric sequence. The length of the radius of the first circle is 6cm. The length of the radius of each of the circles is 4/5 of the length of the radius of the previous circle. Show that the total area of all the circles formed in this way is 100pi cm2.

FuNk

The radius of the first circle is $6$ cm, the second $6\times \frac{4}{5}$ cm, and the radius of the $n$th circle
is $6 \times \left( \frac{4}{5} \right)^{n-1}$.

So the areas of these circles are:

$\pi\ 6^2$, $\pi\ 6^2\left(\frac{4}{5}\right)^2$ and $\pi\ 6^2 \left( \frac{4}{5} \right)^{2(n-1)}$.

So the total area of the circles:

$
A=\sum_{n=1}^{\infty}\pi\ 6^2 \left( \frac{4}{5} \right)^{2(n-1)}=\pi\ 6^2 \sum_{n=1}^{\infty} \left( \left( \frac{4}{5} \right)^2\right) ^{n-1}
$

Now the summation in the last expression above is a geometric series and its sum
is:

$
\sum_{n=1}^{\infty} \left( \left( \frac{4}{5} \right)^2\right) ^{n-1}=\frac{1}{1-(\frac{4}{5})^2}=\frac{25}{9}
$
,

and so:

$A=\pi 6^2 \frac{25}{9}=100 \pi$

RonL
• April 12th 2006, 07:09 AM
FuNkY14
Yo thank you very much!