I had a whole answer and my tab crashed. Hate that.
We'll use the known sum of the first n natural numbers,
[tex]\displaystyle \sum_{k=1}^{n} k = \dfrac{n(n+1)}{2}[/tex]
Let a be the first number, the one we seek.
[tex]\displaystyle 3^9= \sum_{k=0}^{26} (a+2k) = 27a+ 2 \sum_{k=1}^{26}k[/tex]
[tex]\displaystyle 3^9=27a+2(26)(26+1)/2 [/tex]
[tex]3^9=27(a+26) [/tex]
[tex]3^6= a+ 26[/tex]
[tex]a = 3^6 - 26[/tex]
[tex]a = 703 \quad\checkmark[/tex]
