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(B) The $4$-digit numbers divisible by $7$ form an arithmetic progression: $1001, 1008, 1015, \ldots, 9996$.Here,the first term $a = 1001$ and the com

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$5498.5$

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(B) The $4$-digit numbers divisible by $7$ form an arithmetic progression: $1001, 1008, 1015, \ldots, 9996$.Here,the first term $a = 1001$ and the common difference $d = 7$.The last term $l = a + (n-1)d$,so $9996 = 1001 + (n-1)7$.$8995 = (n-1)7 \implies n-1 = 1285 \implies n = 1286$.Since the number of terms $n = 1286$ is even,the median is the average of the $\left(\frac{n}{2}ight)^{	ext{th}}$ and $\left(\frac{n}{2}+1ight)^{	ext{th}}$ terms.Median $= \frac{a_{643} + a_{644}}{2}$.$a_{643} = 1001 + (643-1)7 = 1001 + 4494 = 5495$.$a_{644} = 1001 + (644-1)7 = 1001 + 4501 = 5502$.Median $= \frac{5495 + 5502}{2} = \frac{10997}{2} = 5498.5$.

The median of all $4$-digit numbers that are divisible by $7$ is

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