Question

Let $A_1,A_2,...A_{n}and{H}_1,H_2,...H_n$ are $n$ $A.Ms$ & $H.Ms$ between the two quantities $a$ & $b$ $(a,b>0)$ respectively. On the basis is of above information answer the following questionsThe value of $\frac{A_1}{H_1}-1$ is equal to

• A. $\frac{n}{{(n-1)}^2}{(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}})}^2$
• B. $\frac{n}{{(n-1)}^2}{(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}})}^2$
• C. $\frac{n}{{(n+1)}^2}{(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}})}^2$
• D. $\frac{n}{{(n+1)}^2}{(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}})}^2$

Answer 1

The correct option is D $\frac{n}{{(n+1)}^2}{(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}})}^2$

Given that, $a,A_1,A_2,.....,A_n,b$ are in A.P
Therefore, $\frac{b-a}{n+2-1}=A_1-a$
$\Rightarrow A_1=\frac{an+b}{n+1}$ ...(1)
and $a,H_1,H_2,.....,H_n,b$ are in H.P
Therefore, $\frac{1}{a}.\frac{1}{H_1},\frac{1}{H_2},....,\frac{1}{H_n},\frac{1}{b}$ are in A.P

$\Rightarrow \frac{1}{n+2-1}(\frac{1}{b}-\frac{1}{a})=\frac{1}{H_1}-\frac{1}{a}$
$\frac{1}{H_1}=\frac{a+bn}{ab(n+1)}$ ...(2)

From (1) & (2), we get

$\frac{A_1}{H_1}-1=\frac{(an+b)(a+bn)}{ab{(n+1)}^2}-1=\frac{n}{{(n+1)}^2}(\frac{a}{b}+\frac{b}{a}-2)$
Therefore, $\frac{A_1}{H_1}-1=\frac{n}{{(n+1)}^2}{(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}})}^2$

Ans: D