Question

(i) 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 HP, then find $a$ and $b$.
(ii) If $x,y,z$ are in HP, then show that $log(x+z)+log(x+z-2y)=2log(x-z)$

Answer 1

Now, $a+b=(a+x+y+z+b)-(x+y+z)$
$=\frac{5}{2}(a+b)-15$
[since, $a,x,y,z$ are in AP]
$\therefore$ Sum $=\frac{5}{2}(a+b)\Rightarrow a+b=\mathrm{10...}(i)$
Since, $a,x,y,z,b$ are in HP, then $\frac{1}{a},\frac{1}{x},\frac{1}{y},\frac{1}{z},\frac{1}{b}$
are in AP.
Now, $\frac{1}{a}+\frac{1}{b}=(\frac{1}{a}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{b})-(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$
$=\frac{5}{2}(\frac{1}{a}+\frac{1}{b})-\frac{5}{3}$
$\Rightarrow \frac{a+b}{ab}=\frac{10}{9}\Rightarrow ab=\frac{9\times 10}{10}$ [from Eq. (i)]
$\Rightarrow ab=\mathrm{9...}(ii)$
On solving Eqs. (i) and (ii), we get
$a=1,b=9$
$(ii)LHS=log(x+z)+log(x+z-2y)$
$=log(x+z)+log[x+z-2(\frac{2xz}{x+z}\left)\right][\because y=\frac{2xz}{x+z}]$
$=log(x+z)+log\frac{(x-z{)}^2}{(x+z)}$
$=2log(x-z)+log=RHS$