Let the positive numbers a_1, a_2, a_3, a_4 and a_5 be in a G.P.

Question

Let the positive numbers $a_1, a_2, a_3, a_4$ and $a_5$ be in a G.P. Let their mean and variance be $\dfrac{31}{10}$ and $\dfrac{m}{n}$ respectively, where m and n are co-prime. If the mean of their reciprocal is $\dfrac{31}{40}$ and $a_3+a_4+a_5=14$, then $m+n$ is equal to \_\_\_\_\_\_.

Answer: 211

Accepted Answer

Let terms be $\dfrac{a}{r^2},\dfrac{a}{r},a,ar,ar^2$.

$$\frac{a}{r^2}+\frac{a}{r}+a+ar+ar^2 = 5\times\frac{31}{10} \quad \text{...(1)}$$
$$\frac{r^2}{a}+\frac{r}{a}+\frac{1}{a}+\frac{1}{ar}+\frac{1}{ar^2} = 5\times\frac{31}{40} \quad \text{...(2)}$$

Dividing (1) by (2): $a^2=4 \implies a=2$. From (1): $r+\dfrac1r = \dfrac52 \implies r=2$ (taking $r\ne-2$).

Terms: $\dfrac12, 1, 2, 4, 8$.

$$\sigma^2 = \frac{\sum x^2}{5}-\left(\frac{\sum x}{5}\right)^2 = \frac{186}{25}$$

Hmm --- using the given mean $31/10$: $\sigma^2 = \dfrac{(1/4+1+4+16+64)}{5}-\left(\dfrac{31}{10}\right)^2 = \dfrac{85.25}{5}-9.61 = 17.05-9.61=7.44=\dfrac{186}{25}$

$m+n = 186+25 = 211$

Solutions · 1