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Q.
If $a, b, c$ are in H.P., then the value of $\frac{b+a}{b-a}+\frac{b+c}{b-c}$ is always equal to
Sequences and Series
Solution:
Clearly, $\frac{1}{ a }, \frac{1}{ b }, \frac{1}{ c }$ are in A.P.
Let common difference of A.P. be d.
$\therefore \text { Given expression }= \frac{\frac{1}{ a }+\frac{1}{ b }}{\frac{1}{ a }-\frac{1}{ b }}+\frac{\frac{1}{ c }+\frac{1}{ b }}{\frac{1}{ c }-\frac{1}{ b }}=\frac{\frac{1}{ a }+\frac{1}{ b }}{- d }+\frac{\frac{1}{ c }+\frac{1}{ b }}{ d } $
$ =\frac{-\frac{1}{ a }-\frac{1}{ b }+\frac{1}{ c }+\frac{1}{ b }}{ d }=\frac{\frac{1}{ c }-\frac{1}{ a }}{ d }=\frac{2 d }{ d }=2$