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If a,b,c are in H.P. then value of (1/b+1/c-1/a)(1/c+1/a-1/b),is:-?

If a,b,c are in H.P. then value of (1/b+1/c-1/a)(1/c+1/a-1/b),is:-?

Grade:11

4 Answers

Arun
25750 Points
6 years ago
a, b, c ............. are in H.P. 
1/a, 1/b, 1/c ..... are in A.P. 
1/a + 1/c = 2/b ......................... (1) 
 
the given expression is 

= ( 1/a + 1/b - 1/c )( 1/b + 1/c - 1/a ) 

= [ 1/b + ( 1/a - 1/c ) ] · [ 1/b - ( 1/a - 1/c ) ] 

= (1/b)² - ( 1/a - 1/c )² 

= 1/b² - [ ( 1/a² + 1/c² ) - 2/ac ] 

= 1/b² - { [ ( 1/a + 1/c )² - 2/ac ] - 2/ac } ..... Note This Step 

= 1/b² - [ ( 1/a + 1/c )² - 4/ac ] 

= 1/b² - [ ( 2/b )² - 4/ac ] ................ from (1) 

= 1/b² - 4/b² + 4/ac 

= 4/ac - 3/b² 
Adarsh
15 Points
5 years ago
 
a, b, c ............. are in H.P. 
1/a, 1/b, 1/c ..... are in A.P. 
1/a + 1/c = 2/b ......................... (1)
1/a2+1/c2+2/ac=4/b2
1/a2+1/c2=4/b2-2/ac...................... (2)
 
the given expression is 

= ( 1/a + 1/b - 1/c )( 1/b + 1/c - 1/a ) 

= [ 1/b + ( 1/a - 1/c ) ] · [ 1/b - ( 1/a - 1/c ) ] 

= (1/b)² - ( 1/a - 1/c )² 

= 1/b² - [ ( 1/a² + 1/c² ) - 2/ac ] 

= 1/b² -  [ ( 4/b2-2/ac )- 2/ac ]   ..........from eq(2)

= 1/b² -[4/b2-4/ac]

= 1/b² - 4/b² + 4/ac 

= 4/ac - 3/b² 
tatti
13 Points
3 years ago
mofo stfu iit shit and it’s a total waste of your 2 best years so best enjoy it dumbass iit mera lund bc 
Vikas TU
14149 Points
3 years ago
Dear student 
Please do not misuse the forum ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

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