If a , b , c are three terms of an arithmetic progression such that a is not equal to b , then a − b b − c is always
Take a , b , c as real numbers.
Clarifications: a , b and c may not be consecutive.
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Let us assume the three terms of A.P ( a , b , c ) as ( x , x + d , x + 2 d ) where d = 0 .
a − b b − c = x − ( x + 2 d ) ( x + d ) − ( x + 2 d ) = − d − d = 1
1 is a rational,integer,complex, natural number.
What's wrong in my solution?
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It's not given a,b,c are consecutive AP terms And complex number is a superset of rational number
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Please add the statement that a,b,c may not be consecutive terms. It becomes more clear to the readers.
But a rational number must also be a complex number which is purely real.. Isn't it?
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@Rishabh Jain – Rational numbers are ratios of two integers (for example 3 1 , 1 0 0 1 1 2 3 , 3 4 , . . . ). Integers are rational numbers too. Complex numbers involve a real part and imaginary part. A complex number z can be expressed as z = x + y i where x and y are real, i = − 1 (the imaginary unit), and the real and imaginary parts are x and y respectively. Read more about it here .
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@Chew-Seong Cheong – Note that the complex numbers include all the real numbers (with b = 0 in a + b i above)- That's clearly written on the page you mentioned... And rational numbers are real only.. I guess :-)
@Chew-Seong Cheong – Indeed we classify Complex number a + i b as purely real, purely imaginary or imaginary if b = 0 , a = 0 , b = 0 respectively. Real numbers fall under the first category..
This information that a,b, and c are not consecutive must be given beforehand.
If a,b,and are in ap then , b-c=md where m is +ve integer a-b =nd where n is +ve integer so b-a/a-b=m/n m/n is rational...
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If a , b and c are three terms of an AP , then we can write:
⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ b − c = m d where m = { positive integer if b > c negative integer if b < c a − b = n d where n = { positive integer if a > b negative integer if a < b and d = the common different > 0
Therefore a − b b − c = n d m d = n m a rational number.