Box 1 contains three cards bearing numbers 1, 2, 3; Box 2 contains five cards bearing numbers 1, 2, 3, 4, 5; and Box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let $x_i$ be the number on the card drawn from the $i^{th}$ box, $i=1,2,3$

What is the probability that $x_1,x_2,x_3$ are in A.P. (Arithmetic progression)?

**
Note
**
:- If
$a,b,c$
are in A.P.,
$2b=a+c$

$\dfrac {10}{105}$
$\dfrac {12}{105}$
$\dfrac {11}{105}$
$\dfrac {9}{105}$

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Using the note, we consider all possible values of $x_2$ and count the number of choices of $x_1$ and $x_3$ for which $x_1+x_3 = 2x_2$

Thus there are $1+3+3+3+1 = 11$ possible arithmetic progressions that can be formed. There are $3\times 5\times 7 = 105$ total sequences, so the probability of drawing an arithmetic sequence is $\boxed{\dfrac{11}{105}}$ .