Sequences and Series 5 Question 9
9.
(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 $\log (x+z)+\log (x+z-2 y)=2 \log (x-z)$.
(1978, 3M)
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Answer:
Correct Answer: 9.(i) $a=1, b=9$
Solution:
- (i) 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]
$\therefore \quad \operatorname{Sum}=\frac{5}{2}(a+b) \Rightarrow a+b=10$
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} \quad \frac{1}{a}+\frac{1}{b}-\frac{5}{3} $
$\Rightarrow \quad \frac{a+b}{a b}=\frac{10}{9} \Rightarrow a b=\frac{9 \times 10}{10} \quad$ [from Eq. (i)]
$\Rightarrow \quad a b=9$
On solving Eqs. (i) and (ii), we get
$ a=1, b=9 $
(ii) $LHS=\log (x+z)+\log (x+z-2 y)$
$ \begin{aligned} & =\log (x+z)+\log x+z-2 \frac{2 x z}{x+z} \quad \because y=\frac{2 x z}{x+z} \\ & =\log (x+z)+\log \frac{(x-z)^{2}}{(x+z)} \\ & =2 \log (x-z)=\text { RHS } \end{aligned} $
