Question

The value of$x+y+x$ is $15,$if $a,x,y,z$ and bare in AP while the value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is $\frac{5}{3}$,If,$\frac{1}{a},\frac{1}{x},\frac{1}{y},\frac{1}{z}$ and $\frac{1}{b}$ are in AP . Find the values of a and b.

Answer 1

We have , $x+y+z=15...\left(i\right)$

and $a,x,y,z$and b are in AP.

$\therefore a+x+y+z+b=\frac{5}{2}(a+b)$

$[\therefore S_n=\frac{n}{2}(a_1+a_n\left)\right]$

$\Rightarrow a+15+b=\frac{5}{2}(a+b)$ [from Eq.(i).]

$\Rightarrow a+b=10...\left(ii\right)$

similarly, $\frac{1}{a},\frac{1}{x},\frac{1}{y},\frac{1}{z}and.\frac{1}{b}$are in AP.

$\therefore \frac{1}{a}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{b}=\frac{5}{2}(\frac{1}{a}+\frac{1}{b})$

$\Rightarrow \frac{1}{a}+\frac{5}{3}+\frac{1}{b}=\frac{5}{2}(\frac{1}{a}+\frac{1}{b})$

$[\because \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{5}{3}]$

$\Rightarrow \frac{1}{a}+\frac{1}{b}=\frac{10}{9}$

On solving Eqs. (ii) and (iii) ,we get a=1 and b=9.