1/b-c + 1/b-a = 1/a + 1/c
1/b-c + 1/b-a = 1/a + 1/c
shubham mishra , 13 Years ago
Grade 11
Algebra> prove this .. if a, b , c, are in h.p....
3 AnswersLatika Leekha
I am not able to interpret your query. Please write down the problem clearly so that I can provide you the meaningful answer.
Thanks & Regards
Latika Leekha
askIITians faculty
SHAIK AASIF AHAMED
Hello student,
a, b, c are in HP, so 1/a, 1/b, 1/c are in AP,
->2/b - 1/a = 1/c
->(2a - b) / ab = 1/c
->c = ab/(2a - b)
=1/(b-a) + 1/(b-c)
->2/b - 1/a = 1/c
->(2a - b) / ab = 1/c
->c = ab/(2a - b)
=1/(b-a) + 1/(b-c)
= 1/(b-a) + 1/(b - ab/(2a - b) )
= 1/(b-a) + (2a - b)/( b(2a - b) - ab)
= 1/(b-a) + (2a - b)/ { b(2a - b - a) }
= 1/(b-a) + (2a - b) / {b ( a - b) }
= 1/(b-a) - (2a - b) / {b(b-a) }
= (b - (2a - b) )/ ( b(b-a) )
= 1/(b-a) + (2a - b)/( b(2a - b) - ab)
= 1/(b-a) + (2a - b)/ { b(2a - b - a) }
= 1/(b-a) + (2a - b) / {b ( a - b) }
= 1/(b-a) - (2a - b) / {b(b-a) }
= (b - (2a - b) )/ ( b(b-a) )
=(2b - 2a)/(b(b-a))
= 2(b-a)/(b(b-a)) = 2/b=1/a+1/c
Hence (1/b-c)+(1/b-a)=1/a+1/c
Thanks and Regards
Shaik Aasif
askIITians faculty
PARTH
Hello student,
a, b, c are in HP, so 1/a, 1/b, 1/c are in AP,
->2/b - 1/a = 1/c
->(2a - b) / ab = 1/c
->c = ab/(2a - b)
=1/(b-a) + 1/(b-c)
->2/b - 1/a = 1/c
->(2a - b) / ab = 1/c
->c = ab/(2a - b)
=1/(b-a) + 1/(b-c)
= 1/(b-a) + 1/(b - ab/(2a - b) )
= 1/(b-a) + (2a - b)/( b(2a - b) - ab)
= 1/(b-a) + (2a - b)/ { b(2a - b - a) }
= 1/(b-a) + (2a - b) / {b ( a - b) }
= 1/(b-a) - (2a - b) / {b(b-a) }
= (b - (2a - b) )/ ( b(b-a) )
= 1/(b-a) + (2a - b)/( b(2a - b) - ab)
= 1/(b-a) + (2a - b)/ { b(2a - b - a) }
= 1/(b-a) + (2a - b) / {b ( a - b) }
= 1/(b-a) - (2a - b) / {b(b-a) }
= (b - (2a - b) )/ ( b(b-a) )
=(2b - 2a)/(b(b-a))
= 2(b-a)/(b(b-a)) = 2/b=1/a+1/c
Hence (1/b-c)+(1/b-a)=1/a+1/c
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