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Explanation:

Given numbers are a1, a2, a3,..., a51,100.

If the arithmetic mean of these 52 numbers is ‘x’, we have
a1 + a2 + a3 + ... + a51 + 100 = 52x   ...(1)

Given: the arithmetic mean of a2, a3, a4,...,a51, 100 is ‘x + 1’
a2 + a3 + ⋯ + a51 + 100 = 51(x + 1) = 51x + 51   ...(2)

Solving (1) and (2), we get:
∴ a1 = x − 51   ...(3)

Now for a1 to be maximum possible, x has to be maximum possible.

For x to be maximum possible each of a2, a3, ..., till a51 has to be maximum possible. But they are also distinct integers.
∴ Maximum value of
a51 = 99 [it has to be an integer less than a52]
a50 = 98 [it has to be an integer less than a51]
​​​​​​​...
a2 = 50

⇒ a2, a3, ..., a52 are consecutinve integers from 50 till 100.

⇒ Average of (a2, a3, ..., a52) = 1/2 × (50 + 100) = 75 = x + 1 [From (2)]
⇒ x = 74

From (3)
⇒ a1 = 74 - 51 = 23

Hence, option (b).

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