Question

If a^2(b+c),b^2(c+a),c^2(a+b) are in AP..
Prove that either a,b,c are in A.P or ab+bc+ca=0

Answer 1

$\huge\mathfrak\green{Heyaa!!}$

$\huge\mathfrak\red{Answer:-}$

A, b, c are in A.P.

⇒b-a=c-b

⇒2b=c+a..........(i)

(i)Now,

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

⇒ b2(c+a)-a2(b+c)= c2(a+b)- b2(c+a)

⇒ b2c+ b2a-a2b- a2c= c2a+ c2b- b2c-a b2

⇒ b2c+ b2c+ b2a+a b2-a2b- a2c-c2a- c2b=0

⇒ 2b2c-a2b+2a b2- c2b=ac(c+a)

⇒ ba(2b-a)+bc(2b-c)=ac(c+a)

⇒ bac+bca=ac(c+a) [From(i)]

⇒ 2abc=ac(c+a)

⇒ 2b=c+a

⇒ 2b=c+a which is true.

Hence, a, b, c are in A.P.

$\huge\mathfrak\purple{Hope it helps!!!}$

Answer 2

||✪✪ QUESTION ✪✪||

If a^2(b+c),b^2(c+a),c^2(a+b) are in AP..

Prove that either a,b,c are in A.P or ab+bc+ca=0 ?

|| ✰✰ ANSWER ✰✰ ||

Given That, a²*(b+c),b²*(c+a),c²*(a+b) are in AP..

So,

Common Difference b/w them is same ..

→ [b²(c+a)] - [a²(b+c)] = [c²(a+b)] - [b²(c+a)]

→ (b²c+b²a) - (a²b+a²c) = (c²a+c²b) - (b²c+b²a)

→ b²c - a²c + b²a - a²b = c²a - b²a + c²b - b²c

→ c(b² - a²) + ab(b - a) = a(c² - b²) + bc(c - b)

using (x² - y²) = (x+y)(x-y) now,

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

Taking common now,

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

→ (b - a)[ cb + ca + ab] = (c - b)[ ac + ab + bc]

→ (b - a)[cb + ca + ab] - (c - b)[ ac + ab + bc] = 0

→ [cb + ca + ab] * [ (b - a) - ( c - b) ] = 0

Now, Either [cb + ca + ab] = 0 ,

Or, [ (b - a) - ( c - b) ] = 0

→ [cb + ca + ab] = 0 = Proved.

Or,

→ (b - a) - ( c - b) = 0

→ ( b - a) = ( c - b)

→ 2b = a + c .

Hence, a, b and C are in AP. = Proved.

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