if a,b,c are in AP
Then prove that a (1/b + 1/c ) , b (1/c + 1/a ), c (1/a + 1/b ) are in AP

Question

if a,b,c are in AP
Then prove that a (1/b + 1/c ) , b (1/c + 1/a ), c (1/a + 1/b ) are
in AP

Vijay Kumar Gupta · Accepted Answer

Dear student,

Please find below the solution to the asked query:

Since a, b and c are in AP
 This gives    b-a=c-b   
1Consider the following terms
   a1b+1c, b1c+1a , c1a+1bFind the difference of cosecutive terms
   b1c+1a-a1b+1c=bc+ba-ab-ac                                                =ab2+cb2-a2c-a2babc                                                =ab2-a2b+cb2-a2cabc                                                =abb-a+cb2-a2abc                                                =abb-a+cb-ab+aabc                                                =b-aab+cb+aabc                                                =b-aab+bc+caabcand     c1a+1b-b1c+1a=ca+cb-bc+ba                                                =bc2+ac2-ab2-cb2abc                                                =bc2-cb2+ac2-ab2abc                                                 =bcc-b+ac2-b2abc                                                =bcc-b+ac-bc+babc                                                =c-bbc+ac+babc                                                =c-bab+bc+caabcNote that both are difference have same terms except b-a  and c-bBut this is already equal using 1This shows that difference between the consecutive terms is same
This proves that a1b+1c, b1c+1a , c1a+1b are in AP

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if a,b,c are in AP Then prove that a (1/b + 1/c ) , b (1/c + 1/a ), c (1/a + 1/b ) are in AP

if a,b,c are in AP
Then prove that a (1/b + 1/c ) , b (1/c + 1/a ), c (1/a + 1/b ) are in AP