Question

If 1/a,1/b,1/c are in A.P. prove that a(b+c),b(c+a),c(a+b) are in A.P.​

Answer 1

Answer:

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) —

• Now , we should write this as –

• Now subtract 1 from all terms –

(Hence proved)

Step-by-step explanation:

Answer 2

$\huge\bold{\mathtt{\purple{A {\pink {N {\green {S{\blue{W {\red{E {\orange {R}}}}}}}}}}}}} \\ \\ \bf \: given : \frac{1}{a} . \frac{1}{b} . \frac{1}{c} are \: in \: a.p \\ \therefore \: \frac{ 2}{b} = \frac{1}{a} + \frac{1}{c} \\ \bf \: = > 2ac + ab + bc......(1) \\ to \: prove : a(b+c),b(c+a),c(a+b) are in \: A.P. \\ = > 2b(c + a) = a(b + c) + c(a + b) \\ \bf \: { \bold{lhs}} = 2b(c + a) \\ = > 2bc + 2ba \\ \sf \: \bf \: rhs = a(b + c) + c(a + b) \\ \: = > ab + ac + ac + bc \\ = > ab + 2ac + bc \\ = > ab + ab + bc + bc(from \: (1)) \\ = > 2ab + 2bc \\ \\ \bf \therefore \: lhs = rhs \\ hence \: proved$