Sequences and Series 5 Question 18

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:

  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]

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

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

$$ \begin{aligned} & =\log (x+z)+\log x+z-2 \frac{2 x z}{x+z} \quad \because y=\frac{2 x z}{x+z} \\ & =\log (x+z)+\log \frac{(x-z)^{2}}{(x+z)} \\ & =2 \log (x-z)=\text { RHS } \end{aligned} $$

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