Question

If a^2 (b+ c), b^2 (c+ a), c^2(a+b) are in A.P., then prove that --

(A) a,b,c are in A.P.
(B) ab+bc+ca = 0​

Answer 1

Answer:

• Since, a²(b+c), b²(c+a), c²(a+b) ... are in A.P.
• Since, a²(b+c), b²(c+a), c²(a+b) ... are in A.P.So, b²(c+a) - a²(b+c) = c²(a+b) - b²(c+a)
• Since, a²(b+c), b²(c+a), c²(a+b) ... are in A.P.So, b²(c+a) - a²(b+c) = c²(a+b) - b²(c+a)==> b²c + b²a - a²b - a²c = c²a + c²b - b²c - b²a
• Since, a²(b+c), b²(c+a), c²(a+b) ... are in A.P.So, 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)
• Since, a²(b+c), b²(c+a), c²(a+b) ... are in A.P.So, 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)
• Since, a²(b+c), b²(c+a), c²(a+b) ... are in A.P.So, 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)==>(b-a) [ c(b+a) + ab ] = (c-b) [ a(c+b) + bc ]
• Since, a²(b+c), b²(c+a), c²(a+b) ... are in A.P.So, 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)==>(b-a) [ c(b+a) + ab ] = (c-b) [ a(c+b) + bc ]==> (b-a)( ab + bc ca ) = (c-b)( ab + bc + ca )
• Since, a²(b+c), b²(c+a), c²(a+b) ... are in A.P.So, 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)==>(b-a) [ c(b+a) + ab ] = (c-b) [ a(c+b) + bc ]==> (b-a)( ab + bc ca ) = (c-b)( ab + bc + ca )∴ Either : ab + bc + ca = 0
• Since, a²(b+c), b²(c+a), c²(a+b) ... are in A.P.So, 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)==>(b-a) [ c(b+a) + ab ] = (c-b) [ a(c+b) + bc ]==> (b-a)( ab + bc ca ) = (c-b)( ab + bc + ca )∴ Either : ab + bc + ca = 0Or : b - a = c - b, i.e., a, b, c are in A.P.

Answer 2

Step-by-step explanation:

(A)2b=a+c

divide by bc

2/c=a/bc+1/b

a/bc,1/c,1/b

Hence proved

(B)bc=-ca-ab

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

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

ca=-b(c+a)--------(iv)

=1/a^-bc+1/b^-ca+1/c^-ab

from equation (i),(ii)&(iii)

=1/a^+a(b+c)+1/b^+b(a+c)+1/c^+c(a+b)

=1/a(a+b+c)+1/b(a+b+c)+1/c(a+b+c)

=ab+bc+ca/abc(a+b+c)

ab+bc+ca=0

Hence proved