5 members of a team are weighed consecutively | vedclass

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(C) Let the weight of the first member be $w_1$.The average weight of the first member is $A_1 = w_1$.Let the weight of the second member be $w_2$. Th

Vedclass · Accepted Answer

$8$

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(C) Let the weight of the first member be $w_1$.The average weight of the first member is $A_1 = w_1$.Let the weight of the second member be $w_2$. The average weight of the first two members is $A_2 = \frac{w_1 + w_2}{2} = A_1 + 1 = w_1 + 1$.Thus,$w_1 + w_2 = 2w_1 + 2$,which implies $w_2 = w_1 + 2$.For the $n$-th member,the average weight $A_n = A_{n-1} + 1$.The sum of weights of $n$ members is $S_n = n 	imes A_n$.Since $A_n = A_1 + (n-1) = w_1 + n - 1$,we have $S_n = n(w_1 + n - 1)$.The weight of the $n$-th member is $w_n = S_n - S_{n-1}$.$w_n = n(w_1 + n - 1) - (n-1)(w_1 + n - 2) = nw_1 + n^2 - n - (nw_1 + n^2 - 2n - w_1 - n + 2) = w_1 + 2n - 2$.For the first member $(n=1)$: $w_1 = w_1 + 2(1) - 2 = w_1$.For the fifth member $(n=5)$: $w_5 = w_1 + 2(5) - 2 = w_1 + 8$.The difference between the last and the first player is $w_5 - w_1 = (w_1 + 8) - w_1 = 8 \ kg$.

$5$ members of a team are weighed consecutively and their average weight is calculated after each member is weighed. If the average weight increases by $1 \ kg$ each time,how much heavier is the last player than the first one? (in $kg$)

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