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 $\frac{A_1{H}_1}{ab}$ is equal to

• A. $\frac{a-nb}{b-na}$
• B. $\frac{a+bn}{b+an}$
• C. $\frac{an-b}{bn-a}$
• D. $\frac{b+na}{a+nb}$

Answer 1

The correct option is D $\frac{b+na}{a+nb}$

$a,A_1,A_2,...,A_n,b$ are in AP

$b=a+(n+1)d$

$\Rightarrow d=\frac{b-a}{n+1}$

$A_1=a+d=a+\frac{b-a}{n+1}=\frac{na+b}{n+1}$

$\frac{1}{a},\frac{1}{A_1},\frac{1}{A_2},...,\frac{1}{b}$ are in AP

$\Rightarrow \frac{1}{b}=\frac{1}{a}+(n+1)d$

$\Rightarrow d=\frac{a-b}{ab(n+1)}$

$\therefore \frac{1}{H_1}=\frac{1}{a}+d=\frac{1}{a}+\frac{a-b}{ab(n+1)}$

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

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

$A_1{H}_1=\frac{na+b}{n+1}\frac{ab(n+1)}{a+bn}$

$\frac{A_1{H}_1}{ab}$$=\frac{na+b}{a+bn}$