Question

If a,b,c are in A.P., prove that b+c, c+a, a+b are also in A.P.​

Answer 1

Answer:

Hence (b+c), (c+a) and (a+b) are in A.P

Step-by-step explanation:

Given that-

a, b and c are in A.P.

We know that-

If a, b and c are in A.P then-

$b = \dfrac{a+c}{2}$

2b = a+c

If (b+c), (c+a) and (a+b) are in A.P then-

$(c+a) = \dfrac{(b+c)+(a+b)}{2}$

               = $\dfrac{b+c+a+b}{2}$

               = $\dfrac{a+2b+c}{2}$

But 2b = a+c

Then, $c+a = \dfrac{a+c+a+c}{2}$

                       = $\dfrac{2a+2c}{2}$

                       = $\dfrac{2(a+c)}{2}$

a + c = a + c

L.H.S = R.H.S

Hence (b+c), (c+a) and (a+b) are in A.P

Answer 2

Answer:

a , b , c are in ap So , 2b = a+c For b+c , c+a , a+b to be in ap u need to show that here also 2b= a+c 2(a+c) = b+c+a+b a+c = 2b Done..... Hope u understood , and 2b= a+c is like an identity , u can call it in ap !