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If a2,b2,c2\xa0are in A.P to prove that 1/b+c,1/c+a,1/a+b are in A.P |
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Answer» If a2, b2, c2\xa0are in AP, then2b2\xa0= a2\xa0+ c2\xa0b2\xa0+ b2\xa0= a2\xa0+ c2\xa0b2\xa0- a2\xa0= c2\xa0- b2\xa0(b - a)(b + a) = (c - b)(c + b)(b - a)/(c + b) = (c - b)/(b + a)Dividing both sides by (c + a),(b - a)/{(c + b)(c + a)} = (c - b)/{(b + a)(c + a)}{(b + c) - (c + a)}/{(b + c)(c + a) = {(c + a) - (a + b)}/{(a + b)(c + a)}{1/(c + a)} - {1/(b + c)} = {1/(a + b)} - {1/(c + a)}{2/(c + a)} = {1/(a + b)} - {1/(b + c)}Therefore 1/(a + b), 1/(b + c), 1/(c + a) are in AP. Ans. Given : a2 b2 c2 are in A.P.To Prove :\xa0\\({1 \\over b +c } ,{1 \\over c+a}, {1\\over a+b} \\)\xa0Are in A.P.Proof :\xa02b2 = a2 + c2 [as a2 b2 c2 are in A.P.]=> b2 + b2 = a2 + c2=> b2 - a2 = c2 - b2\xa0=> (b-a) (b+a) = (c-b) (c+b)=>\xa0\\({(b-a) \\over (c+b) } = {(c-b) \\over (b+a)}\\)Divide both side by\xa0\\(1\\over (c+a)\\), We get\xa0=>\xa0\\({(b-a) \\over (c+b) \\times (c+a) } = {(c-b) \\over (b+a)\\times(c+a)}\\)=>\xa0\\({(b-a+c-c) \\over (c+b) \\times (c+a) } = {(c-b+a-a) \\over (b+a)\\times(c+a)}\\)=>\xa0\\({(b+c) - (c+a) \\over (c+b) \\times (c+a) } = {(c+a) -(a+b)) \\over (b+a)\\times(c+a)}\\)=>\xa0\\({1 \\over(c+a)} - {1\\over (c+b) } = {1\\over (a+b)} -{ 1\\over(c+a)}\\)=>\xa0\\({2 \\over(c+a)} = {1\\over (a+b)} + {1\\over (c+b) }\\)Hence by this equation we, can say that\xa0\\({1 \\over b +c } ,{1 \\over c+a}, {1\\over a+b} \\)are in A.P. |
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