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 false, statement 2 is true.
B. Statement 1 is true, statement 2 is true, statement 2 is a correct explanation for statement 1.
C. Statement 1 is true, statement 2 is true, statement 2 is not a correct explanation for statement 2.
D. Statement 1 is true, statement 2 is false.

Answer

D

Explanation

Total number of ways of selecting four numbers = `.^(20)C_(4)` Let d be common difference of A.P.
If d = 1, we have
(1, 2, 3, 4), (2, 3, 4, 5), …, (17, 18, 19, 20) `rarr` 17 A.P.s
If d = 2, we have
(1, 3, 5, 7), (2, 4, 6, 8), …, (14, 16, 18, 20) `rarr` 14 A. P.s
If d = 3, we have
(1, 4, 7, 10), (2, 5, 8, 11), ...,(11, 14, 17, 20) `rarr` 11 A.P.s
If d = 4, we have
(1, 5, 9, 13), (2, 6, 10, 14), ..., (8, 12, 16, 20) `rarr` 8 A.P.s
If d = 5, we have
(1, 6, 11, 16), (2, 7, 12, 17), ..., (5, 10, 15, 20) `rarr` 5 A.P.s
If d = 6, we have
(1, 7, 13, 19), (2, 8, 14, 20) `rarr` 2 A.P.s
So required probability
`=(17+14+11+8+5+2)/(.^(20)C_(4))=(1)/(85)`