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Probability Level 2

If ( 23 r − 1 ) 23\choose r-1 , ( 23 r ) 23\choose r and ( 23 r + 1 ) 23\choose r+1 are in arithmetic progression , then which of the following cannot be a value of r r ?

This is an original problem and belongs to the set My creations

9 None of these 14 17

Skanda Prasad by Skanda Prasad , 20, India

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1 solution

Skanda Prasad
Sep 22, 2017

If ( n r − 1 ) n\choose r-1 , ( n r ) n\choose r and ( n r + 1 ) n\choose r+1 are in A P AP then n n and r r must satisfy the condition given by

( n − 2 r ) 2 = n + 2 (n-2r)^2=n+2 .

Substituting n = 23 n=23 , we have ( 23 − 2 r ) 2 = 23 + 2 (23-2r)^2=23+2

⟹ \implies 23 − 2 r = 5 23-2r=5 and 23 − 2 r = − 5 23-2r=-5 .

⟹ \implies r = 9 r=9 and r = 14 r=14

Hence, r ≠ 17 r\neq 17 .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Why must ( n − 2 r ) 2 = n + 2 (n-2r)^2 = n + 2 be satisfied?

Pi Han Goh - 3 years, 8 months ago

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Actually even I'm not sure... I found this formula in a solution of a probability problem.

To tell you the truth, I was thinking of posting a note to ask about the proof for this...

Skanda Prasad - 3 years, 8 months ago

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Hint: If a , b , c a,b,c follows an arithmetic progression , then b − a = c − b b-a = c-b .

Pi Han Goh - 3 years, 8 months ago

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