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$