Let a1, a2, ... , a52 be positive integers such that a1

Question

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

A. 20
B. 23
C. 45
D. 48
E. 50

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lacktutor · Accepted Answer

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 \frac{(a_1 + a_2 +...+ a_{52})}{52} +1 = \frac{(a_2 +a _3+...+ a_{52})}{51 } \frac{(a_1 + a_2 +...+ a_{52}+ 52)}{52}= \frac{(a_2 +a _3+...+ a_{52})}{51} 51 (a_1 + a_2 +...+ a_{52}+ 52) =52 (a_2 +a _3+...+ a_{52}) --> 51a _1 + 51*52= a_2 +a _3+...+ a_{52} a_{52} =100 In order a _1 to be the largest possible value, a_2 +a _3+...+ a_{52} = 50 +51 +52 +...+100 = \frac{(100+50)*51}{2}= 75*51 --> 51a _1 + 51*52= 75*51 a _1= 75 -52 = 23 Answer (B).

chetan2u · Answer

Sometimes understanding can make a question so simple..

Now, the crux is .. 
You remove  a_1 , and you get the average of remaining 51 numbers down by 1.
Thus  this number has to be 51+1(itself) less than the average.

If we want  a_1  to be the maximum, the average of remaining 51 should be the highest, and for that all numbers should be maximum possible and that is they should be consecutive. 
So  a_2=a_{52}-51+1=100-51+1=50 ..
Average of consecutive numbers =  \frac{50+100}{2}=75 .

a_1=75-(51=1)=75-52=23

B

varun325 · Answer

chetan2u  CAN YOU EXPLAIN THE CRUX "You remove a1, and you get the average of remaining 51 numbers down by 1.
Thus this number has to be 51+1(itself) less than the average."

ScottTargetTestPrep · Answer

Let S = a2 + a3 + … + a51 + 100, i.e., the sum of the 52 terms except the first term. We can create the equation:

(a1 + S)/52 + 1 = S/51

(a1 + S)/52 + 52/52 = S/51

(a1 + S + 52)/52 = S/51

51(a1 + S + 52) = 52S

51a1 + 51S + 2652 = 52S

51a1 = S - 2652

a1 = S/51 - 52

The largest value of S is a2 + a3 + … + a51 + 100 = 50 + 51 + … + 99 + 100 = (50 + 100)/2 x 51 = 75 x 51 = 75 x 51. So, if S = 75 x 51, then a1 = S/51 - 52 = 75 - 52 = 23 would be the largest possible value.

Answer: B

chetan2u · Answer

Look say average of 51 numbers is a, so sum is 51*a. But say average of 52 numbers is a-1 now, that is 52(a-1)=52a-52. 
As you can see if 52nd number was a, the average would remain as a and sum as 52a.

But the new average is 52 less than 52a, that is it is same as removing 1 from each of the 52 number. 
This is basically visualising the question and at many times this can make a question very simple and avoid complex calculations.

rudi0setyawan · Answer

For a1 can be max, a2 until a51 must be maximum too
So a2 will be 100 - 50 = 50
And the mean will be (100 + 50)/2 = 75

If we just continue the sequence by giving a1 = 49, the mean will be (100 + 49)/2 = 74 .5 
If we want to get rid this  0.5 , simply deduct 49 by ( 0.5 *52)
= 49 - 26
= 23

Kinshook · Answer

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

\frac{a_1 + a_2 + ... + a_{52}}{52} = \frac{ a_2 + ... + a_{52}}{51} - 1  
 51*(a_1 + a_2 + ... + a_{52}) = 52*(a_2 + ... + a_{52}) - 51*52  
 51*a_1 = a_2 + ... + a_{52} - 51*52

For largest value of a_1; all values from a_2,...a_{52} should be largest possible

51*a_1 = 50 + ... +100 + 2652 = 101*100/2 - 49*50/2 - 51*52 = 1173  
 a_1 = \frac{1173}{51} = 23

IMO B

Let a1, a2, ... , a52 be positive integers such that a1 < a2 < ... < a

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