Question

If 1/a , 1/b , 1/c are in AP. Then PT. b+c/a , c+a/b, a+b/c are in AP​

Answer 1

GIVEN :–

Terms 1/a , 1/b , 1/c are in A.P.

TO PROVE :–

(b+c)/a , (c+a)/b , (a+b)/c are in A.P.

SOLUTION :–

• According to the question –

=> 1/a , 1/b , 1/c are in A.P.

• Now multiply with (a + b + c) —

$\\ \implies \: \frac{(a + b + c)}{a} \: , \: \frac{a + b + c}{b} \: , \: \frac{a + b + c}{c} \: \: are \: \: in \: \: A.P.$

• Now , we should write this as –

$\\ \implies \: \frac{(b + c)}{a} + 1 \: , \: \frac{(c + a)}{b} + 1 \: , \: \frac{(a + b)}{c} + 1 \: \: are \: \: in \: \: A.P. \:$

• Now subtract 1 from all terms –

$\\ \implies \: \frac{b + c}{a} \: , \: \frac{c + a}{b} \: , \: \frac{a + b }{c} \: \: are \: \: in \: \: A.P. \:$

(Hence proved)