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Question 2 : Let a_{1}, a_{2}, ... , a_{52} be positive integers such that a_{1} ＜ a_{2} ＜ ... ＜ a_{52}. Suppose, their arithmetic mean is one less than the arithmetic mean of a_{2}, a_{3}, ..., a_{52}. If a_{52} = 100, then the largest possible value of a_{1} is

- 48
- 20
- 45
- 23

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It is given that a_{1},a_{2},......,a_{52} are positive integers such that a_{1} < a_{2} < ....... < a_{52} and it is also given that their arithmetic mean is one less than the arithmetic mean of a_{2},......,a_{52}. Let us consider that a_{2},......,a_{52} have the arithmetic mean of M and from a_{1} to a_{52} the arithmetic mean is M - 1.

If we add a_{1} to a_{2},......,a_{52}, the average falls down by 1 or the total falls by 52. So, a_{1} = M - 52

In conventional way if sum of a_{2} + a_{3} + .... + a_{52} = 51M. Sum when a_{1} is added = 51M + a_{1}. Average of a_{1},a_{2},......,a_{52} = M – 1.

\\frac{51M + a_1}{52}) = M - 1

51M + a_{1} = 52M – 52

a_{1} = M - 52

We have to find the maximum value of a_{1} and this value can be maximum when average or mean M tends to be maximum. We know that all the numbers are distinct i.e from a_{2},......,a_{52} there are 51 numbers. The maximum possible average of 51 numbers if the largest number is 100.

The numbers are i.e 50,51,.....,100. Average = \\frac{100 + 50}{2}). Hence, the maximum possible average is 75.

M_{max} = 75

A_{max} = 75 - 52

A_{max} = 23

The largest possible value of a_{1} is 23

The question is **"Let a _{1}, a_{2}, ... , a_{52} be positive integers such that a_{1} ＜ a_{2} ＜ ... ＜ a_{52}. Suppose, their arithmetic mean is one less than the arithmetic mean of a_{2}, a_{3}, ..., a_{52}. If a_{52} = 100, then the largest possible value of a_{1} is "**

Choice D is the correct answer.

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