Question

if a,b,c are in AP, show that
`[(b+c)^(2) -a^(2)],[(c+a)^(2) -b^(2)],[(a+b)^(2) =c^(2)]` are in AP.

Answer 1

Since a,b,c in AP, we have
2b =( a+c)
Now, ` [(b +c)^(2) -a^(2) ], [( c+a)^(2) -b^(2) ],[(a+b)^(2) -c^(2)] `will be in AP .
` if [(c +a)^(2) -b^(2)] - [(b+c)^(2) -a^(2)] =[(a+b)^(2) -c^(2)]-[(c +a)^(2) -b^(2)]`
i.e, ` if ( a+b+c) ( c+a-b) -( a+b+c) ( b+c-a) `
= ` ( a+b+c) ( a+b-c) -( a +b +c) ( c+a-b) `
i.e, ` if ( a+b+c) [( c+a-b) -( b+c-a)]`
= ( a+b+c) [( a+b -c) -( c +a -b)]
i.e, if 2(a-b) = 2 ( b-c)
i.e, if ( a-b) = ( b-c)
i.e if 2b = ( a+c) , which is true by (i).
a,b,c, are in AP.
` Rightarrow [(b+c)^(2) -a^(2)] [(c+a)^(2) -=b^(2)], [(a +b)^(2) -c^(2)]` are in AP.