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 $$\frac{31}{10}$$ and $$\frac{m}{n}$$ respectively, where $m$ and $n$

Accepted Answer

Since ${a}_{1},{a}_{2},{a}_{3},{a}_{4},{a}_{5}$ are in geometric progression, we can write :

${a}_{2}=ra$,

${a}_{3}={r}^{2}a$,

${a}_{4}={r}^{3}a$,

${a}_{5}={r}^{4}a$.

where $r$ is the common ratio and ${a}_{1}$ = $a$ is the first term.

Given that the mean of the series is $\frac{31}{10}$, we get

$\frac{1}{5}({a}_{1}+{a}_{2}+{a}_{3}+{a}_{4}+{a}_{5})=\frac{31}{10}$

Substituting the terms with the values of ${a}_{1}$ and $r$ gives

$\frac{1}{5}(a+ra+{r}^{2}a+{r}^{3}a+{r}^{4}a)=\frac{31}{10}$

Simplifying this gives

$a(1+r+{r}^{2}+{r}^{3}+{r}^{4})=\frac{31}{2}$

$\begin{gathered} \frac{a({r}^{5}-1)}{r-1}=\frac{31}{2}......(1) \ \frac{1}{a}(1+\frac{1}{r}+\frac{1}{{r}^{2}}+\frac{1}{{r}^{3}}+\frac{1}{{r}^{4}})=\frac{31}{40}\cdot 5=\frac{31}{8} \ \frac{1}{a}(\frac{1-{(\frac{1}{r})}^{5}}{1-\frac{1}{r}})=\frac{31}{8} \ \frac{1}{a}(\frac{{r}^{5}-1}{r-1})\frac{1}{{r}^{4}}=\frac{31}{8}.........(2) \end{gathered}$

From (1) and (2)

$\frac{1}{a}\cdot \frac{31}{2a}\cdot \frac{1}{{r}^{4}}=\frac{31}{8}$

$a{r}^{2}=2$

From (1)

$\begin{gathered} \frac{2}{{r}^{2}}(\frac{{r}^{5}-1}{r-1})=\frac{31}{2} \ \frac{1+r+{r}^{2}+{r}^{3}+{r}^{4}}{{r}^{2}}=\frac{31}{4} \ ({r}^{2}+\frac{1}{{r}^{2}})+(r+\frac{1}{r})=\frac{27}{4} \ {t}^{2}-2+t=\frac{27}{4} \end{gathered}$

$\begin{gathered} 4{t}^{2}+4t-35=0 \ 4{t}^{2}+14t-10t-35=0 \ (2t-5)(2t+7)=0 \ t=\frac{5}{2},\frac{-7}{2}\Rightarrow r=2 \ 	herefore r=2,a=\frac{1}{2} \end{gathered}$

${\frac{1}{2},1,2,4,8}$

$	herefore {\sigma }^{2}=\frac{\sum {X}^{2}}{N}-{(\frac{\sum X}{N})}^{2}$

$\begin{gathered} =\frac{(\frac{341}{4})}{5}-{(\frac{31}{10})}^{2} \ =\frac{341}{20}-\frac{961}{100}=\frac{1705-961}{100} \ =\frac{744}{100} \end{gathered}$

$=\frac{186}{25}=\frac{m}{n}\Rightarrow 211=m+n$

Solutions · 1