Question

if a, b, c are in AP then prove that a square bracket of B + c, b square bracket of C plus a, c square bracket of a + b are also in ap ap

Answer 1

a , b , c are in AP
so, common difference must be constant.
b - a = c - b
now, multiply with (ab + bc + ca) in both sides,
(b - a)(ab + bc + ca) = (c - b)(ab + bc + ca)
(b - a)[ab + c(b + a)] = (c - b)[bc + a(c + b)]
(b - a)ab + c(b + a)(b - a) = (c - b)bc + a(c + b)(c - b)
(ab² - a²b) + c(b² - a²) = (c²b - bc²) + a(c² - b²)
(ab² - a²b) + (cb² - a²c) = (c²b - b²c) + (ac² - ab²)
(ab² + cb²) -(a²b + a²c) = (c²b + ac²) - (bc² + ac²)
$\boxed{\boxed{ {b}^{2} (a + c) - {a}^{2} (b + c) = {c}^{2} (b + a) - {b}^{2} (c + a)}}$

you can observed that ,
it seems like a²(b + c) , b²(c + a), c²(a + b) are in AP

Answer 2

Answer:

answer is yes

Explanation:

a , b , c are in AP

so, common difference must be constant.

b - a = c - b

now, multiply with (ab + bc + ca) in both sides,

(b - a)(ab + bc + ca) = (c - b)(ab + bc + ca)

(b - a)[ab + c(b + a)] = (c - b)[bc + a(c + b)]

(b - a)ab + c(b + a)(b - a) = (c - b)bc + a(c + b)(c - b)

(ab² - a²b) + c(b² - a²) = (c²b - bc²) + a(c² - b²)

(ab² - a²b) + (cb² - a²c) = (c²b - b²c) + (ac² - ab²)

(ab² + cb²) -(a²b + a²c) = (c²b + ac²) - (bc² + ac²)

\boxed{\boxed{ {b}^{2} (a + c) - {a}^{2} (b + c) = {c}^{2} (b + a) - {b}^{2} (c + a)}}

b

2

(a+c)−a

2

(b+c)=c

2

(b+a)−b

2

(c+a)

you can observed that ,

it seems like a²(b + c) , b²(c + a), c²(a + b) are in AP