If H$$_1$$ , H$$_2$$ , H$$_3$$ , ..., H$$_n$$ , are 'n' Harmonic means between ‘a’ and ‘b’ (≠ a), then value of $$\dfrac{H_{1}+a}{H_{1}-a}+\dfrac{H_{n}+b}{H_{n}-b}$$ is equal to

Let us assume that n = 3 and a, b = 2, 6. 

Therefore, the harmonic sequence will be: 2, H$$_1$$, H$$_2$$, H$$_3$$, 6

Hence, H$$_2$$ = $$\dfrac{2*2*6}{2+6}$$ = 3

H$$_1$$ = $$\dfrac{2*2*3}{2+3}$$ = $$\dfrac{12}{5} = 2.4$$

H$$_3$$ = $$\dfrac{2*3*6}{3+6}$$ = $$4$$

Therefore, $$\dfrac{H_{1}+a}{H_{1}-a}+\dfrac{H_{n}+b}{H_{n}-b}$$

$$\dfrac{2.4+2}{2.4-2}+\dfrac{4+6}{4-6}$$

$$\Rightarrow$$ $$11-5$$ = 6. 

Option B: 2n = 2*3 = 6.