Four numbers are chosen at random (without replacement) from the set {1, 2, 3, ....., 20}. Statement-1: The probability that the chosen numbers when arranged in some order will form an AP Is `1/(85)` . Statement-2: If the four chosen numbers form an AP, then the set of all possible values of common difference is {1, 2, 3, 4, 5}. (1) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is false (3) Statement-1 is false, Statement-2 is true (4) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Answer

C

Explanation

Total number of ways of selecting 4 numbers out of first 20 natural numbers = ` .^(20)C_(4)` .
Number of A.P with common differenece 1 is =17
(1 ,2,3,4,, 2,3,4,5 ,….., , 17 ,18,19,20)
Number of A.P's with common differen 2 is =14
( 1,3,5,7,, 2,4,6,8,..., 14,16,18,20)
number of A,P s with common difference 2 is 14
Number of A.P's with common differen 3 is =11
( 1,4,7,10, 2,5,8,11 ,...., 11,14,17,20) and soon.
The maximum common difference can be 6 and the number of such A.P 's is = 2
( 1,7,13,19, 2,8, 14,20)
Total number of ways in which chosen numbers are in A.p
17 + 14 +11 + .....+ 2=57
Propbability than the chosen numbers are in AP
`57/(""^(20)C_(4))=1/85`
Clearly, the set of all possible values of common difference is
` {+-1,+-2,+-3,+-4,+-5,+-6}`
So, statement -2 is not true.