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

1.	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» 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.

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»

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.

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

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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​

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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