Question

The value of $x+y+z=15$, if $a,x,y,z,b$ are in A.P, while the value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{5}{3}$ if $a,x,y,z,b$ are in H.P. Find $a\times$ $b$.

Answer 1

To find the value of $a$ and $b$

Given,

$x+y+z=15$....(i)

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{5}{3}$........(ii)

if $a,x,y,z$ are in A.P

Then, $x,y,z$ are in A.P

$\therefore 2y=x+z$......(iii)

$\Rightarrow 2y+y=15$........... comparing (iii) and (i)

$\Rightarrow 3y=15\Rightarrow y=5$

Now,

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

$\Rightarrow a+b=10$......(vi)

Now, if $a,x,y,z$ are in HP

then $x,y,z$ one in H.P

$\therefore \frac{2}{y}=\frac{1}{x}+\frac{1}{z}$.......(iv)

$\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{5}{3}$.......(v)

from (iv) and (v) we can write

$\frac{2}{y}+\frac{1}{y}=\frac{5}{3}=\frac{5}{3}$........putting (iv) in (v)

$\frac{3}{y}=\frac{5}{3}\Rightarrow y=\frac{9}{5}$

Also, $a,y,b$ are in H.P.

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

$\Rightarrow \frac{a+b}{ab}=\frac{-2}{y}=\frac{2\times 5}{9}$

$\Rightarrow \frac{a+b}{ab}=\frac{10}{9}$......(vi)

by comparing (vi) and (vii) we can say that

$\frac{a+b}{ab}=\frac{10}{9}$

$\Rightarrow \frac{10}{ab}=\frac{10}{9}$

$\Rightarrow ab=9$.......(viii)

$\Rightarrow a(10-a)=9$

$\Rightarrow 10a-a^2=9\Rightarrow a^2-10a+9=0$

$\Rightarrow (a-1)(a-9)=0$

$\therefore a=1$ or $9$

$\therefore$ when $a=1,b=9$

$a=9,b=1$

$\therefore a\times b=9\times 1=1\times 9=9$