Question

Two distinct, real, infinite geometric series each have a sum of 1 and have the same second term. The third term of one of the series is $1/8$. If the second term of both the series can be written in the form $\frac{\sqrt{m}-n}{p}$ where $m,n$ and $p$ are positive integers and $m$ is not divisible by the square of any prime, find the value of $100m+10n+p$

Answer 1

Answer: 518

Let $a_1,r_1$ be the first term and the common ratio of the first series.
hence $\frac{a_1}{1-r_1}=1\Rightarrow a_1+r_1=1$
also $a_1{r}_1^2=\frac{1}{8}\Rightarrow a_1=\frac{1}{8r_1^2}$
substituting in (1)
$\frac{1}{8r_1^2}+r_1=1\Rightarrow 8r_1^3-8r_1^2+1=0$
$4r_1^2\left(2r_1-1\right)-2r_1\left(2r_1-1\right)-1\left(2r_1-1\right)=0$
$\left(2r_1-1\right)\left(4r_1^2-2r_1-1\right)=0$
hence $r_1=\frac{1}{2}$ or $r_1=\frac{2\pm \sqrt{4+16}}{8}=\frac{1\pm \sqrt{5}}{4}$
$\therefore r_1=\frac{1}{2}\text{ or }{r}_1=\frac{1+\sqrt{5}}{4}\text{ or }{r}_1=\frac{1-\sqrt{5}}{4}$
corresponding values of $1^{\text{st }}$ term are,
$a_1=\frac{1}{2}\text{ or }{a}_1=\frac{3-\sqrt{5}}{4}\text{ or }{a}_1=\frac{3+\sqrt{5}}{4}\text{ [using }{a}_1+r_1=1]$
$\text{ now }{2}^{\text{nd }}\text{ term }{a}_1{r}_1=\frac{1}{4}\text{ or }\frac{\sqrt{5}-1}{8}\text{ or }\frac{-\sqrt{5}-1}{8}$
only the $2^{\text{nd }}$ is of the given form
hence $m=5;n=1;p=8$
$\therefore 100m+10n+p=518$