Two adjacent terms of an arithmetic sequence differ by a fixed amount - the common difference.
In this case, the terms differ by 7. So, we can write your sequence as
[tex]\begin{array}{ccc}a_1&12&12+0\cdot 7\\a_2&19&12+1\cdot 7\\a_3&26&12+2\cdot 7\\a_4&33\\\vdots&\vdots&\vdots\\a_n&110&12+(n-1)\cdot 7\end{array}[/tex]
So, first of all, we can observe that the last term is
[tex]12+7(n-1)=110 \iff 7(n-1)=98 \iff n-1=14 \iff n=15[/tex]
So, we're summing the first 15 terms of the sequence:
[tex]\displaystyle \sum_{n=1}^{15} 12+7(n-1) = \sum_{n=1}^{15} 12 + \sum_{n=1}^{15} 7(n-1) = \sum_{n=1}^{15} 12 + 7\sum_{n=1}^{15} n-1[/tex]
We can work out the sums as follows: it is trivial that
[tex]\displaystyle \sum_{n=1}^{15} 12 = 12\cdot 15 = 180[/tex]
As for the other sum, we have
[tex]\displaystyle 7\sum_{n=1}^{15} n-1 = 7\sum_{n=0}^{14}n = 7\dfrac{14\cdot 15}{2} = 735[/tex]
So, the value is
[tex]\displaystyle \sum_{n=1}^{15} 12+7(n-1) = 180+735= 915[/tex]
