Question

prove that if 1/a,1/b,1/care in ap then a(b+c) , b(a+c) , c(a+b) are in ap

Answer 1

Answer:

If 1/a , 1/b , 1/c are in AP, these must satisfy 2*middle term = 1st tern + 3rd term

= > 2(1/b) = 1/a + 1/c

= > 2(1/b) = (a+c)/ac

= > 2ac = (a+c)b

= > 2ac = ab + bc

Adding ab + bc to both sides:

= > 2ac + ab + bc = ab + bc + ab + bc

= > ab + ac + ac + bc = 2( ab + bc )

= > a( b + c ) + c( a + b ) = 2b( a + c )

= > 1st term + 2nd term = 2*middle term

This, new equation, also satisfies the condition of AP, hence this, too, is an AP. Proved that a(b+c), b(a+c), c(a+b) are in AP.