Question

if (b+c)/a, (c+a)/b and (a+b)/c are in AP, show that 1/a, 1/b and 1/c are in AP​

Answer 1

hope its help u..............

Answer 2

$\mathfrak{\Large{\underline{\underline{Solution : }}}}$

$\mathsf{\underline{Given,}} \\ \\ \\ \sf \implies \dfrac{( b \: + \: c )}{a}, \: \dfrac{( c \: + \: a )}{b}, \: \dfrac{ ( a \: + \: b)}{c} \: are \: in \: A.P. \\ \\ \\ \textsf{Add 1 to all terms : } \\ \\ \\ \sf \implies \dfrac{ b \: + \: c }{a} \: + \: 1, \: \dfrac{c \: + \: a }{b} \: + \: 1, \: \dfrac{ a \: + \: b}{c} \: + \: 1 \: are \: in \: A.P.$

$\sf \implies \dfrac{ b \: + \: c \: + \: a }{a} , \: \dfrac{c \: + \: a \: + \: b }{b},\: \dfrac{ a \: + \: b \: + \: c}{c} \ are \: in \: A.P. </p><p></p><p> \\ \\ \\ \sf \implies \dfrac{ \cancel{ a \: + \: b \: + \: c}}{a} , \: \dfrac{ \cancel{a \: + \: b \: + \: c}}{b},\: \dfrac{ \cancel{ a \: + \: b \: + \: c}}{c} \ are \: in \: A.P. </p><p></p><p> \\ \\ \\ \sf \: \: \therefore \: \: \dfrac{1}{a} \: \: , \: \: \dfrac{1}{b} \: \: , \: \: \dfrac{1}{c} \: \: are \: in \: A.P.$

$\boxed{\Large{\underline{ \mathfrak{Proved \: \: !! }}}}</p><p>$