Question 78

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

Solution

Let 'x' be the average of all 52Ā positive integersĀ $$\ a_{1},a_{2}...a_{52}\ $$.

$$a_{1}+a_{2}+a_{3}+...+a_{52}$$ = 52x ... (1)

Therefore, average ofĀ $$a_{2}$$, $$a_{3}$$, ....$$a_{52}$$ = x+1

$$a_{2}+a_{3}+a_{4}+...+a_{52}$$ = 51(x+1)Ā ... (2)

From equation (1) and (2), we can say that

$$a_{1}+51(x+1)$$ =Ā 52x

$$a_{1}$$ = x - 51.Ā 

We have to find out the largest possible value ofĀ $$a_{1}$$.Ā $$a_{1}$$ will be maximum when 'x' is maximum.Ā 

(x+1) is the average of termsĀ $$a_{2}$$, $$a_{3}$$, ....$$a_{52}$$. We know thatĀ $$a_{2}$$ < $$a_{3}$$ < ... <Ā $$a_{52}\ $$ andĀ $$a_{52}$$ = 100.

Therefore, (x+1) will be maximum when each term is maximum possible. IfĀ $$a_{52}$$ = 100, thenĀ $$a_{52}$$ = 99,Ā $$a_{50}$$ = 98 ends so on.

$$a_{2}$$ = 100 + (51-1)*(-1) = 50.

Hence, Ā $$a_{2}+a_{3}+a_{4}+...+a_{52}$$ = 50+51+...+99+100 =Ā 51(x+1)

$$\Rightarrow$$ $$\dfrac{51*(50+100)}{2} =Ā 51(x+1)$$

$$\Rightarrow$$ $$x =Ā 74$$

Therefore,Ā theĀ largest possible value ofĀ $$a_{1}$$ = x - 51 = 74 - 51 = 23.Ā 

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