Question

1. If a, b, c
are in AP, Prove that a² + c ² - 2 b c = 2 a(b-c)​

Answer 1

Step-by-step explanation:

if a, b, c are in AP then 2b = a + c

a2 + c2 - 2bc= a2 + c2 - ( a+c)c

= a2 + c2 -ac-c2

= a2 - ac

= a(a-c)

= a((2b-c)-c)

= a(2b-2c)

= 2a(b-c)

Answer 2

It is proved that $a^{2} +c^{2} -2\ b\ c = 2a( b-c)$.

Step-by-step explanation:

Given:

a, b, c are in AP.

Prove that  $a^{2} +c^{2} -2\ b\ c = 2a( b-c)$.

To  Find:

It is  to prove that $a^{2} +c^{2} -2\ b\ c = 2a( b-c)$.

Formula Used:

If a, b, c are in  Arithmetic Progression (A.P.).

It means $b-a=c-a$

             $2b=a+c$

Solution:

As given a, b and c are in A.P.

$2b=a+c$   ---------- equation no.01

 As given Prove that $a^{2} +c^{2} -2\ b\ c = 2a( b-c)$

 LHS  $=a^{2} +c^{2} -2\ b\ c$

 Putting the value of b from equation no.01.

         $=a^{2} +c^{2} -2\ (\frac{a+c}{2} )\ c$

          $=a^{2} +c^{2} - (ac+c^{2} )$

          $=a^{2} +c^{2} - ac-c^{2}$

          $=a^{2} - ac$

       

 RHS  $=2a(b-c)$

 Putting the value of b from equation no.01.

           $=2a(\frac{a+c}{2} -c)$

           $=2a(\frac{a+c-2c}{2} )$

          $=2a(\frac{a-c}{2} )$

           $=a(a-c)$

           $=a^{2} - ac$

 LHS= RHS

Hence, it is proved that a² + c ² - 2 b c = 2 a(b-c).