Inquisitive Marv's favorite integer

This question can be easily approached without having to sum up all of the terms.

Observe that the average of the $N$ integers, is the same as the average of the integers paired up from the end. Hence, we have

$\frac{(N+1) + (2N)}{2} =80 .$

Solving this gives us $N = \frac{160 - 1 } { 3} = 53.$

Para encontrar a média, precisamos somar todos os termos e dividir pelo número de parcelas n. Observe que a sequencia dada é uma Progressão aritmética, com isso, podemos obtemos a soma dos termos.

$\frac{n(3N+1)}{2n} = 80$

$\frac{(3N+1)}{2}$ = 80

$3N+1 = 160$

$3N = 159$

$N = 53.$

We start with the following equation, given the stipulated information:

$\frac{N + 1 + N + 2 + N + 3 + ... + 2N}{N}$ = 80

∵ there are N integers, rearranging, we have:

$N^{2}$ + 1 + 2 + 3 + ... + N = 80N

We apply the identity:

1 + 2 + 3 + ... + N = $\frac{N(N+1)}{2}$

∴ $N^{2}$ + $\frac{N(N+1)}{2}$ = 80N

∵ N can't be equal to 0, we can divide by N from all terms:

N + $\frac{N+1}{2}$ = 80

The rest is simple manipulation:

2N + N + 1 = 160

3N = 159

N = 53

∴ Marv's favourite integer is 53.

N+1+N+2...2N= N^2 + N(N+1)/2. The total summation divided by N is the average. Therefore, N + (N+1)/2 = 80 which implies 3N = 159 and so N=53

Consider the Arithmetic Progression, N+1, N+2, .., 2N. Obviously, the number of terms is N. Sum of the terms = N/2 (first term + last term) = N / 2 (N+1 + 2N) = N / 2 (3N + 1) Given, average is 80. Therefore, Sum of the terms / N = 80 Therefore, sum of the terms = 80N. This gives, N(3N + 1) = 160N 3N + 1 = 160 3N = 159 N = 53.

We can write the average of the given equation as following :

$\frac{{\left( {n + 1} \right) + \left( {n + 2} \right) + ... + (n + n)}}{n} = 80$

The upper part can also be written as a sum:

$\frac{1}{n}\sum\limits_{k = 1}^n {\left( {n + k} \right)} = 80$

Which we can develop into:

$\frac{1}{n}\left[ {\sum\limits_{k = 1}^n n + \sum\limits_{k = 1}^n k } \right] = 80$

Then we obtain :

$\begin{array}{l} \frac{1}{n}\left[ {{n^2} + \frac{{n\left( {n + 1} \right)}}{2}} \right] = 80\\ 3n + 1 = 160\\ n = 53 \end{array}$

So Marv's favorite integer is 53.

We start with the following equation, given the stipulated information:

N+1+N+2+N+3+...+2NN= 80

∵ there are N integers, rearranging, we have:

N2 + 1 + 2 + 3 + ... + N = 80N

We apply the identity:

1 + 2 + 3 + ... + N = N(N+1)2 ∴ N2 + N(N+1)2 = 80N

∵ N can't be equal to 0, we can divide by N from all terms:

N + N+12 = 80

The rest is simple manipulation:

2N + N + 1 = 160

3N = 159

N = 53

∴ Marv's favourite integer is 53.

Let n be Marv's favorite integer From n+1 to 2n, there are n integers. Then, $n(3n +1)/2$ = $80n$ , So $n$ =53