Please solve this question again and kindly explain how the denominator get reduced

Q 5 If   1 a + b , 1 b + c   and   1 c + a   are   in   A . P . ,   then   which   of   the   following   relations   is   correct ? ( A )   2 b 2 = a 2 + c 2 ( B )   2 a 2 = b 2 + c 2 ( C )   a 2 + b 2 + c 2 = 0 ( D )   2 c 2 = a 2 + b 2 Solution It   is   given   that   1 a + b , 1 b + c   and   1 c + a   are   in   A . P 1 b + c - 1 a + b = 1 c + a - 1 b + c a + b - ( b + c ) ( b + c ) ( a + b ) = b + c - ( c + a ) ( c + a ) ( b + c ) a - c a + b = b - a c + a a - c a + c = b + a b - a a 2 - c 2 = b 2 - a 2 2 a 2 = b 2 + c 2

Dear student
The question involves the concept of A.P. Arithmatic progression,the scope of which goes much beyond the ambit of your syllabus,but stillI am helping you with the calculations.We have,1b+c-1a+b=1c+a-1b+ca+b-b+ca+bb+c=b+c-c+ac+ab+ca+b-b-ca+b=b+c-c-ac+aa-ca+b=b-ac+aa-cc+a=b+ab-aa2-c2=b2-a2   using x2-y2=x+yx-y2a2=b2+c2
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It is already solved
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Yes it is
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It is already solve.
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My friend it will already solve. But what you ask pls tell me fast.
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