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If a, b, c are in H.P. then the value of
b+ab−a+b+cb−c\frac{b + a}{b - a} + \frac{b + c}{b - c} is

Show answer and solution
a, b, c in H.P
⇒1a,1b,1c\Rightarrow \frac{1}{a},\frac{1}{b},\frac{1}{c} in A.P
Let
1a−a,−d\frac{1}{a} - a, - d
1b=a,\frac{1}{b} = a, 1c=a,+d\frac{1}{c} = a, + d.
Now,

b+ab−a+b+cb−c=1a1+1a1−d1a1−1a1−d+1a1+1a1+d1a1−1a1+d\frac{b + a}{b - a} + \frac{b + c}{b - c} = \frac{\frac{1}{a_{1}} + \frac{1}{a_{1} - d}}{\frac{1}{a_{1}} - \frac{1}{a_{1} - d}} + \frac{\frac{1}{a_{1}} + \frac{1}{a_{1} + d}}{\frac{1}{a_{1}} - \frac{1}{a_{1} + d}}
=2a1−d−d+2a1+dd= \frac{2a_{1} - d}{- d} + \frac{2a_{1} + d}{d}
=2dd=2= \frac{2d}{d} = 2

Source: Grass-G / M-06 Mathematics Paper 14 October.docx

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