Sequences And Series Ques 105

(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)

Show Answer

Answer:

Correct Answer: 9.(i) $a=1, b=9$

Solution:

Formula:

HARMONICAL PROGRESSION (H.P.):

  1. (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]

$\therefore \quad \operatorname{Sum}=\frac{5}{2}(a+b) \Rightarrow a+b=10$ $\quad$ …….(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} \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)]

$\Rightarrow \quad a b=9$ $\quad$ …….(ii)

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} $



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