Integers 1, 2, 3, ldots, n (n geq 3) are writ | vedclass

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(C) The sum of the first $n$ integers is given by $S_n = \frac{n(n+1)}{2}$.After erasing the integer $k$,the sum of the remaining $(n-1)$ integers is

Vedclass · Accepted Answer

$47$

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(C) The sum of the first $n$ integers is given by $S_n = \frac{n(n+1)}{2}$.After erasing the integer $k$,the sum of the remaining $(n-1)$ integers is $\frac{n(n+1)}{2} - k$.The average of the remaining numbers is given as $16$,so:$\frac{\frac{n(n+1)}{2} - k}{n-1} = 16$$n(n+1) - 2k = 32(n-1)$$n^2 + n - 2k = 32n - 32$$2k = n^2 - 31n + 32$$k = \frac{n^2 - 31n + 32}{2}$Since $1 $1 From $k From $k > 1$: $n^2 - 31n + 32 > 2 \implies n^2 - 31n + 30 > 0 \implies (n-30)(n-1) > 0$. Since $n \geq 3$,we have $n > 30$.Thus,$n$ must be $31$.Substituting $n = 31$ into the equation for $k$:$k = \frac{31^2 - 31(31) + 32}{2} = \frac{32}{2} = 16$.Therefore,$n + k = 31 + 16 = 47$.

Class	$4-8$	$8-12$	$12-16$	$16-20$
Frequency	$3$	$\lambda$	$4$	$7$

Integers $1, 2, 3, \ldots, n$ $(n \geq 3)$ are written on a blackboard and an integer $k$ $(1 < k < n)$ is erased. The average of the remaining numbers is $16$. Then $n + k$ is

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