Question

2. The sum of a certain infinite geometric series is 20. when all
the terms in the series are squared the sum of the resulting
series is 80. If the first term of the original series is expressed
in lowest terms as (p/q) where p,qe N, then find the value
of (p+q).​

Answer 1

Let $r$ be the common ratio such that $0<r<1.$ Then the sum of the infinite geometric series will be given by,

$\longrightarrow\dfrac{a}{1-r}=20$

$\longrightarrow r=1-\dfrac{a}{20}$

$\longrightarrow 1+r=2-\dfrac{a}{20}\quad\quad\dots(1)$

Now each term in the series is squared, then,

• first term, $a_1'=a'=a^2.$

• second term, $a_2'=a^2r^2.$

• common ratio, $r'=r^2.$

As $0<r<1,$ we get $0<r^2<1$ too. Then the new sum is given by,

$\longrightarrow\dfrac{a'}{1-r'}=80$

$\longrightarrow\dfrac{a^2}{1-r^2}=80$

$\longrightarrow\dfrac{a^2}{(1-r)(1+r)}=20\times4$

$\longrightarrow\dfrac{a}{1-r}\cdot\dfrac{a}{1+r}=\dfrac{4a}{1-r}$

$\longrightarrow\dfrac{a}{1+r}=4$

From (1),

$\longrightarrow\dfrac{a}{2-\dfrac{a}{20}}=4$

$\longrightarrow\dfrac{20a}{40-a}=4$

$\longrightarrow20a=160-4a$

$\longrightarrow a=\dfrac{160}{24}$

$\longrightarrow a=\dfrac{20}{3}$

Now $a$ is in the simplest form, so,

• $p=20$

• $q=3$

Hence,

$\longrightarrow\underline{\underline{p+q=23}}$