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 $log\, (x+z) + log\, (x+z-2y) = 2\, log\, (x-z)$

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 $=\frac{5}{2}(a+b) \Rightarrow a + b = 10 ...(i)$
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, $\frac{1}{a}+\frac{1}{b} = \bigg(\frac{1}{a}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{b}\bigg) - \bigg(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\bigg)$
$=\frac{5}{2}\bigg(\frac{1}{a}+\frac{1}{b}\bigg)-\frac{5}{3}$
$\Rightarrow \frac{a+b}{ab}=\frac{10}{9} \Rightarrow ab = \frac{9 \times 10}{10}$ [from Eq. (i)]
$\Rightarrow ab = 9 ...(ii)$
On solving Eqs. (i) and (ii), we get
$ a = 1, b = 9 $
$(ii)\, LHS = log\, (x+z) + log\, (x+z-2y) $
$= log\, (x+z) + log\, \bigg[x+z-2\bigg(\frac{2\, xz}{x+z}\bigg)\bigg]\bigg[\because y = \frac{2\, xz}{x+z}\bigg]$
$ = log\, (x+z) + log\, \frac{(x-z)^2}{(x+z)} $
$ = 2\, log\, (x-z) + log = RHS $