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sequence and series
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hi umang bhaiya.........the answer is H.P since a+b/1-ab, b, b+c/1-bc are in A.P (b+c) / (1-bc) - b = b- (a+b)/(1-ab) this implies (b+c-b+b2c) / (1-bc) = (b-ab2-a-b) / (1-ab) c(1+b2)/(1-bc) = -(a(1+b2))/ (1-ab) c-abc= -a+abc this implies 2abc=c+a 2b= (c+a)/ ac = 1/a+1/c this again implies 1/a, b , 1/c are in A.P and hence a, 1/b , c are in H.P pls tell me if i am wrong.........  |
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