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H1 H2.......Hn are n  HM  between a and b where a  is not equal to b then the value of [H1+a/H1-a]+[Hn+b/Hn-b]
10 months ago

i) As given terms are in HP, let (1/a), (1/H1), (1/H2), ---- (1/Hn), (1/b) are in AP
[Total there are (n+2) terms and 'n' means]

ii) ==> (1/b) = (1/a) + (n + 1)d, where d is the common difference of AP
Solving d = (a - b)/(ab(n + 1)}

iii) ==> 1/H1 = 1/a + d = (b*n + a)/{ab(n + 1)}
==> H1 = ab(n + 1)/(b*n + a)

==> H1/a = (b*n + b)/(b*n + a)

Applying compenendo & dividendo property of ratios and simplifying,
[If a/b = c/d, then (a+b)/(a-b) = (c+d)/(c-d)],

(H1 + a)/(H1 - a) = (2b*n + a + b)/(b - a)

iv) In similar process, taking 1/Hn = (1/b) - d

we get, (Hn + b)/(Hn - b) = (2a*n + a + b)/(a - b) = -(2a*n + a + b)/(b - a)

v) So from the above, adding (iii) & (iv),

[(H1 + a)/(H1 - a)] + [(Hn + b)/(Hn - b)] = (2b*n + a + b - 2a*n - a - b)/(b - a)

= 2n(b - a)/(b - a) = 2n

Thus it is proved that [(H1 + a)/(H1 - a)] + [(Hn + b)/(Hn - b)] = 2n

10 months ago
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