Question

If a b c are in ap (b+c)² - a² , (c+a)² - b² , (a+b)² - c² are also in ap

Answer 1

Given: a , b , c are in AP

To find: Prove that (b + c)² - a² , (c + a )² - b², (a + b)² - c² are in AP.

Solution:

Now we have given: a , b , c are in AP.

Multiply it by 2, we get:

                2a , 2b , 2c

Now subtraction (a + b + c) from each term, we get:

                2a - (a + b + c ) , 2b -(a + b + c ), 2c - (a + b + c)

Solving it, we get:

                (a - b - c), (b - c - a), (c - a - b)

We can rewrite it as:

                -(b + c - a), -(c + a - b) , -(a + b - c)

                (b + c - a), (c + a - b), (a + b - c )

Now multiplying by (a + b + c) to each term, we get:

                (b + c - a)(b + c + a), (c + a - b)(c + a + b), (a + b - c)(a + b +c )

Now solving it, we get:

                (b + c)² - c², (c + a )² - b², (a + b)² - c² which are in AP.

Hence proved.

Answer:

     So we proved that (b + c)² - a² , (c + a )² - b², (a + b)² - c² are in AP.