Question

If abc are in ap then show that a+ 1/bc, b + 1/ac, c + 1/ab are also in ap

Answer 1

given a, b ,c are in AP
then ,2b=a+c
now abc+1/ac=(abc+1/bc +abc+1/ab)/2
2/ac=1/bc+1/ab
2/ac=a+c/abc
2=a+c/b
2b=a+c
Hence ,proved

Answer 2

a+ 1/bc, b + 1/ac, c + 1/ab

Step-by-step explanation:

$\frac{1}{bc}, \frac{1}{ac}, \frac{1}{ab}$

$\frac{1}{ca} - \frac{1}{bc}$ = $\frac{bc - ca}{abc^{2} }$

= $\frac{c (b - a)}{abc^{2} }$ = $\frac{b-a}{abc}$

$\frac{1}{ab} - \frac{1}{ca} = \frac{1}{a} (\frac{1}{b} - \frac{1}{c}$)

= $\frac{1}{a} (\frac{c - b}{bc} )$

We know that,

a, b, c in A.P.

b - a = c - b

= $\frac{b-a}{abc}$

1/bc, 1/ca, 1/ac in A.P.

a + 1/bc, b + 1/ca, c + 1/ac = A.P

Learn more: AP

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