Grade 11Algebra

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

shubhankit

8 Years agoGrade 11

4 Answers

Arun

8 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

7 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 eqⁿ(2)

= 1/b² -[4/b²-4/ac]

= 1/b² - 4/b² + 4/ac

= 4/ac - 3/b²