Integer sequencing

While the author's solution is correct, this hopefully explains it more fully. Each sequence will be an AP with common difference $d=1$ .

Then for any sequence, let $x$ be the first term and $n$ be the number of terms. The sum of the sequence is $[(x) + (x+n-1)]\frac{n}{2}$ (first term plus last term multiply by the number of pairs; where the last term is $x+n-1$ because the sequence is consecutive integers).

The sum must equal 30, therefore $(2x+n-1)\frac{n}{2} = 30 \rightarrow 2x+n-1 =\frac{60}{n}$ . Since $n$ is always an integer, $\frac{60}{n}$ is an integer only if $n$ divides 60.

If $\frac{60}{n}$ is an integer, $\frac{60}{n} + 1 - n$ must be even for $x$ to be an integer.

The first sequence must be at least 3 terms long.

if $n = 3$
$2x+3-1 = \frac{60}{3} \rightarrow x=9 \rightarrow$ Sequence is 9, 10, 11, LAST TERM = 11

If $n=4$
$2x+4-1 = \frac{60}{4} \rightarrow x=6 \rightarrow$ Sequences is 6, 7, 8, 9, LAST TERM = 9

If $n=5$
$2x+5-1 = \frac{60}{5}\rightarrow x=4 \rightarrow$ Sequence is 4, 5, 6, 7, 8, LAST TERM = 8

If $n=6$
$2x+6-1=\frac{60}{10} \rightarrow 2x=5 \rightarrow x$ is not an integer

If $n=7, 8, 9$ , $\frac{60}{n}$ is not an integer
If $n=10: 2x+9=6$ and $x$ is negative, but the problem specifies that we are using positive integers, therefore $n$ cannot be larger or equal to 10.

Thus the only 3 sequences are in fact the ones above, and the sum of the last terms is $11 + 9 + 8 = 28$ .

There are 3 such sequences:

9, 10, 11

6, 7, 8, 9

4, 5, 6, 7, 8

The final terms are 11, 9, and 8. 11 + 9 + 8 = 28.