Question

Assertion :$H_1,H_2,H_3,....H_n$ be $n$ $H.M$ between $a$ and $b$ then value of $\frac{H_1+a}{H_1-a}+\frac{H_n+b}{H_n-b}=2n$ Reason: $H_1=\frac{(n+1)ab}{nb+a},H_n=\frac{(n+1)ab}{na+b}$ obtained by interchanging the numbers $a$ & $b$.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

$a,H_1,H_2,....H_n,b$ are in $H.P$
$\Rightarrow \frac{1}{a},\frac{1}{H_1}...\frac{1}{H_2},\frac{1}{b}$ are in $A.P$
$\therefore d=\frac{a-b}{ab(n+1)}$
So $\frac{1}{H_1}=t_2=\frac{a+nb}{(n+1)ab}\therefore H_1=\frac{(n+1)ab}{a+nb}$
$\Rightarrow \frac{H_1}{a}=\frac{(n+1)ab}{nb+a}$ .....(A)
& $\Rightarrow H_n=\frac{(n+1)a}{na+b}$ by interchanging $a$ & $b$
$\frac{H_n}{b}=\frac{(n+1)a}{na+b}$ ....(B)

Now using componendo & dividendo on (A) & (B) and then adding, we get
$\frac{H_1+a}{H_1-a}+\frac{H_n+b}{H_n-b}=\frac{(2n+1)b+a}{b-a}+\frac{(2n+1)a+b}{a-b}$
$=\frac{\{(2n+1)b+a\}-\{(2n+1)a+b\}}{b-a}$
$=\frac{(2n+1)(b-a)-(b-a)}{b-a}=\frac{(b-a)[2n+1-1]}{b-a}$
Therefore, $\frac{H_1+a}{H_1-a}+\frac{H_n+b}{H_n-b}=2n$
Ans: A