Question

The value of $x+y+z$ is $15$ if $a,x,y,z,b$ are in AP, while the value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}is\frac{5}{3}$ if $a,x,y,z,b$ are in H.P. find $|a-b|$.

Answer 1

We know that sum of n A.M.'s between two quantities is equal to $n$ times their single mean.
Now $x,y,z$ are three A.M.'s between $a$ and $b$
$x+y+x=3\left(\frac{a+b}{2}\right)=15$
or $a+b=10...\left(1\right)$
a,x,y,z,b are in H.P.
$\therefore \frac{1}{a},\frac{1}{x},\frac{1}{y},\frac{1}{z},\frac{1}{b}$ are in A.P.
$\therefore \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{2}(\frac{1}{a}+\frac{1}{b})=3\left(\frac{a+b}{2ab}\right)$
or $\frac{5}{3}=\frac{3}{2ab}.10,$ by (1) $\therefore ab=9$ ...(2)
Hence $a$ and $b$ are the roots of
$t^2-10t+9=0,$ by (1) and (2)
$\therefore t=9,1$ are the required values of $a$ and $b$.