Question

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

• A. Both (A) & (R) are individually true & (R) is correct explanation of (A).
• B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).
• C. (A) is true but (R) is false.
• D. (A) is false but (R) is true

Answer 1

Answer: (A)

$a,H_1,H_2,....H_n,b$ are in $H.P$
$\Rightarrow \frac{1}{a},\frac{1}{H_1},\frac{....1}{H_n},\frac{1}{b}$ are in $A.P$.
$\therefore d=\frac{a-b}{ab\left(n+1\right)}$
So, $\frac{1}{H_1}=t_2=\frac{a+nb}{\left(n+1\right)ab}\therefore H_1=\frac{\left(n+1\right)ab}{a+nb}$
$\Rightarrow \frac{H_1}{a}=\frac{\left(n+1\right)b}{nb+a}...\left(A\right)$
$\&H_n=\frac{\left(n+1\right)ab}{na+b}$
(By interchanging a & b)
$\Rightarrow \frac{H_n}{b}=\frac{\left(n+1\right)a}{na+b}....\left(B\right)$
Now, using componendo & dividendo on (A) & (B) then adding, we get
$\frac{H_1+a}{H_1-a}+\frac{H_n+b}{H_n-b}=\frac{\left(2n+1\right)b+a}{b-a}+\frac{\left(2n+1\right)a+b}{a-b}$
$=\frac{\left\{\left(2n+1\right)b+a\right\}-\left\{\left(2n+1\right)a+b\right\}}{b-a}$
$=\frac{\left(2n+1\right)\left(b-a\right)-\left(b-a\right)}{b-a}$
$=\frac{\left(b-a\right)\left[2n+1-1\right]}{b-a}=2n$