Question

The value of XYZ is $\frac{15}{2}$ if a,x,y,z,b are In A.P., while xyz is $\frac{18}{5}$ if a x,y,x,b are in H.P If a and b are positive integers, then find them.

Answer 1

We have,

Given that,

$xyz=\frac{15}{2}$ if $x,y,z$ in an A.P.

$xyz=\frac{18}{5}$ if $x,y,z$ in H.P.

If $d$ is the common difference this A.P.

$a,x,y,z,b$

Then,

$d=\frac{b-a}{n+1}=\frac{b-a}{4}$

Also $y=\frac{a+b}{2}$

$x=\frac{3a+b}{4}$

And

$z=\frac{3b+a}{4}$

Then,

$xyz=\frac{15}{2}=\left(\frac{3a+b}{2}\right)\left(\frac{a+b}{2}\right)\left(\frac{3b+a}{4}\right)$ …… (1)

If $a,x,y,z,b$ are in H.P.

$\Rightarrow \frac{1}{a},\frac{1}{x},\frac{1}{y},\frac{1}{z},\frac{1}{b}$ are in A.P.

$y=H.M.$ of $aandb$

$\Rightarrow b=y=\frac{2ab}{a+b}$

$\Rightarrow \frac{1}{y}=\frac{a+b}{2ab}$

Now,

$\frac{1}{x}=\frac{\frac{3}{a}+\frac{1}{b}}{4}$

$\Rightarrow \frac{4ab}{3b+a}=x$

$z=\frac{4ab}{3a+b}$

Now, given that,

$xyz=\frac{18}{5}=\left(\frac{4ab}{3b+a}\right)\left(\frac{2ab}{a+b}\right)\left(\frac{4ab}{3a+b}\right)......\left(2\right)$

On multiplying (1) and (2) to and we get,

$a^3{b}^3=\frac{18}{5}\times \frac{15}{2}=27$

${\left(ab\right)}^3=3^3$

Now,

$ab=3$

Then, $aandb$ are positive integer.

Now,

$ab=1\times 3$

$so,$

$a=1,b=3$

Hence, this is the answet.