Question

show that 1/b+c' 1/c+a' 1/a+b are in AP​

Answer 1

$\large\red{ANSWER}$

$\frac{1}{b + c} \: \: \: \frac{1}{c + a} \: \: \: \frac{1}{a + b} \\ \\ a = \frac{1}{b + c} \\ \\ d \: 1 = \frac{1}{c + a} - \frac{1}{b + c} \\ \: \: = \frac{b + c - c - a}{(c + a)(b + c)} \\ \: \: = \frac{b - a}{cb + {c}^{2} + ab + ac } \\ \\ \\ d \: 2 = \frac{1}{a + b} - \frac{1}{c + a} \\ \: \: = \frac{c + a - a - b}{(a + b)(c + a)} \\ \: \: = \frac{c - b}{ac + {a}^{2} + ab + bc }$

If these given 3 numbers are an AP

then,

d 1 =d 2

$\frac{b - a}{cb + {c}^{2} + ab + ac } = \frac{c - b}{ac + {a}^{2} + ab + bc} \\(b - a)(ac + {a}^{2} + ab + bc) = (c - b)(cb + {c}^{2} + ab + ac)$

at the end we will get

$2 + \frac{a}{c} + \frac{c}{a} = 2 + \frac{c}{a} + \frac{a}{c} \\ lhs = \: rhs$

$\large\blue{HENCE\:PROVEN\:THAT\:IT\:IS\:AN\:A.P}$

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