Question

Let $a_1,a_2,a_3{a}_4,a_5$ be a $G.P$ of positive real numbers such that the $A.M$ of $a_2$ and $a_4$ is $117$ and the $G.M$ of $a_2$ and $a_4$ is $108$. Then the $A.M$ of $a_1$ and $a_5$ is:

• A. $108$
• B. $117$
• C. $144.5$
• D. $145.5$

Answer 1

Answer: (D)

$145.5$

Given that $a_1,a_2,a_3,a_4,a_5$ are in G.P.

Also given $a_2+a_4=234$ ...(1)

and $a_2{a}_4=(108{)}^2$ ...(2)

Let $a_2=a_{1}r,a_3=a_1{r}^2,a_4=a_1{r}^3,a_5=a_1{r}^4$

$\therefore a_{1}r(1+r^2)=234$ ..(3)

$a_1^2{r}^4=(108{)}^2\Rightarrow a_1{r}^2=108$ ...(4)

Now lets divide equation (3) by equation (4), we get

$\frac{(1+r^2)}{r}=\frac{234}{108}=\frac{117}{54}=\frac{39}{18}=\frac{13}{6}$

Therefore, $6+6r^2=13r$

$\Rightarrow 6r^2-13r+6=0$

$\Rightarrow (2r-3)(3r-2)=0$

$\Rightarrow r=\frac{3}{2}$ or $\frac{2}{3}$

Since $a_1{r}^2=108\to a_1\times \frac{9}{4}=108$ or $a_1=48$

Thus $a_5=a_1{r}^4=48\times \frac{81}{16}=243$

Therefore, $AM$ of $a_1$ & $a_5=\frac{48+243}{2}=\frac{291}{2}=145.5$

Note: The answer will come same, if we substitute value of $r=\frac{2}{3}$.