Question

THE SECOND QUESTION

THIS QUESTION IS OF 100 POINTS SO PLEASE DO ME A FAVOUR ANSWER IT PROPERLY

Answer 1

Hi...

1). If a,b,c are in AP then 2b=a+c
We have to prove
2(a+c)=(b+c)+(a+b)
so 2a+2c=2b+a+c
so a+c=2b

Now..

2). Given, a,b and c are in A.P.

a2(b + c), b2(c + a), c2(a + b) will be in A.P.

if b2(c + a) – a2(b + c) = c2(a + b) – b2(c + a)

i.e., if c(b2 – a2) + ab(b – a) = a(c2 – b2) + bc(c – b)

i.e., if (b – a)(ab + bc + ca) = (c – b)(ab + bc + ca)

i.e., if b – a = c – b

i.e., if 2b = a + c

i.e., if a, b, c are in A.P.

Thus, if a, b, c are in A.P. then a2(b + c), b2(c + a), c2(a + b) will also be in A.P.




Hope it helps you:-)

Answer 2

If a, b, c are in AP, prove that b + c, c + a, and a + b are also in AP.

The conventional approach is the simplest. As a, b, c are in AP, we can say that 2b = a + c. a, b, c are in AP