Let $$\ a_{1},a_{2}...a_{52}\ $$ be positive integers such that $$\ a_{1}$$ < $$a_{2}$$ < ... <Â $$a_{52}\ $$. 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
Let 'x' be the average of all 52 positive integers $$\ a_{1},a_{2}...a_{52}\ $$.
$$a_{1}+a_{2}+a_{3}+...+a_{52}$$ = 52x ... (1)
Therefore, average of $$a_{2}$$, $$a_{3}$$, ....$$a_{52}$$ = x+1
$$a_{2}+a_{3}+a_{4}+...+a_{52}$$ = 51(x+1)Â ... (2)
From equation (1) and (2), we can say that
$$a_{1}+51(x+1)$$ =Â 52x
$$a_{1}$$ = x - 51.Â
We have to find out the largest possible value of $$a_{1}$$. $$a_{1}$$ will be maximum when 'x' is maximum.Â
(x+1) is the average of terms $$a_{2}$$, $$a_{3}$$, ....$$a_{52}$$. We know that $$a_{2}$$ < $$a_{3}$$ < ... < $$a_{52}\ $$ and $$a_{52}$$ = 100.
Therefore, (x+1) will be maximum when each term is maximum possible. If $$a_{52}$$ = 100, then $$a_{52}$$ = 99, $$a_{50}$$ = 98 ends so on.
$$a_{2}$$ = 100 + (51-1)*(-1) = 50.
Hence, Â $$a_{2}+a_{3}+a_{4}+...+a_{52}$$ = 50+51+...+99+100 =Â 51(x+1)
$$\Rightarrow$$ $$\dfrac{51*(50+100)}{2} =Â 51(x+1)$$
$$\Rightarrow$$ $$x =Â 74$$
Therefore, the largest possible value of $$a_{1}$$ = x - 51 = 74 - 51 = 23.Â
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