A Pre-RMO question! -30

Let the first term of the geometric progression be $a$ and its common ratio be $r$ , where $r<1$ . Then we have:

$\begin{aligned} \frac a{1-r} & =2005 &... (1) \\ \frac {a^2} {1 - r^2} & =20050 \\ \frac a {1 - r}\cdot \frac {a} {1 + r} & =20050 \\ 2005 \cdot \frac a{1 +r} & =20050 \\ \frac a{1 +r} & =10 &... (2) \\ \frac {(1)} {(2)}: \ \frac {1 +r} {1 - r} & = \frac{2005} {10} =\frac{401}2 \\ 2-2r&=401+401r \\ \implies r&= \frac {399}{403} \end{aligned}$

Therefore $m+n=399+403=802$ and its sum of digits is $\boxed {1 0}$ .

Let $\text{first term } = a$ , $\text{common ratio }= r$ and $S = 2005$

$a + ar + ar^2 \ldots \infty = S$ $\boxed{\frac{a}{1-r} = S}\ldots(1)$

$a^2 + a^2r^2 + a^2r^4 \ldots \infty = 10S$ $\boxed{\frac{a^2}{1-r^2} = 10S}\ldots(2)$

$\text{Doing }\displaystyle \frac{(1)^2}{(2)}$

$\frac{1-r^2}{(1-r)^2} = \frac{S}{10}$ $\frac{1+r}{1-r} = \frac{S}{10}$

$\frac{1+r}{1-r} =\frac{2005}{10} = \frac{401}{2}$

Apply Componendo and Dividendo

$\frac{1}{r} = \frac{403}{399}$

$\color{#3D99F6}{\boxed{r = \frac{399}{403}}}$

$m = 399, n = 403, m + n=802 \text{ and Sum of Digits } = 10$