Sequence $(a_n)$ is defined as $a_{n+1}-2a_n+a_{n-1}=7$ for every $n\geq 2$, where $a_1 = 1, a_2=5$. What is $a_{17}$?

$ 
\textbf{(A)}\ 895
\qquad\textbf{(B)}\ 900
\qquad\textbf{(C)}\ 905
\qquad\textbf{(D)}\ 910
\qquad\textbf{(E)}\ \text{None of the above}
$
1. The given recurrence relation is:
   \[
   a_{n+1} - 2a_n + a_{n-1} = 7 \quad \text{for} \quad n \geq 2
   \]
   with initial conditions \(a_1 = 1\) and \(a_2 = 5\).

2. To solve this, we first rewrite the recurrence relation in a more convenient form:
   \[
   (a_{n+1} - a_n) - (a_n - a_{n-1}) = 7
   \]
   Let \(b_n = a_n - a_{n-1}\). Then the recurrence relation becomes:
   \[
   b_{n+1} - b_n = 7
   \]

3. This is a first-order linear difference equation. We solve it by noting that the difference between consecutive terms of \(b_n\) is constant:
   \[
   b_{n+1} = b_n + 7
   \]

4. We need an initial condition for \(b_n\). From the initial conditions of \(a_n\):
   \[
   b_2 = a_2 - a_1 = 5 - 1 = 4
   \]

5. Using the recurrence relation for \(b_n\), we find:
   \[
   b_n = b_2 + 7(n-2) = 4 + 7(n-2)
   \]
   Simplifying, we get:
   \[
   b_n = 7n - 10
   \]

6. Now, we express \(a_n\) in terms of \(b_n\):
   \[
   a_n = a_1 + \sum_{k=2}^n b_k
   \]
   Since \(a_1 = 1\), we have:
   \[
   a_n = 1 + \sum_{k=2}^n (7k - 10)
   \]

7. We split the sum:
   \[
   a_n = 1 + \sum_{k=2}^n 7k - \sum_{k=2}^n 10
   \]

8. The first sum is:
   \[
   \sum_{k=2}^n 7k = 7 \sum_{k=2}^n k = 7 \left( \sum_{k=1}^n k - 1 \right) = 7 \left( \frac{n(n+1)}{2} - 1 \right)
   \]
   Simplifying, we get:
   \[
   7 \left( \frac{n(n+1)}{2} - 1 \right) = \frac{7n(n+1)}{2} - 7
   \]

9. The second sum is:
   \[
   \sum_{k=2}^n 10 = 10(n-1)
   \]

10. Combining these results, we get:
    \[
    a_n = 1 + \left( \frac{7n(n+1)}{2} - 7 \right) - 10(n-1)
    \]
    Simplifying further:
    \[
    a_n = 1 + \frac{7n(n+1)}{2} - 7 - 10n + 10
    \]
    \[
    a_n = 1 + \frac{7n(n+1)}{2} - 10n + 3
    \]
    \[
    a_n = \frac{7n^2 + 7n - 20n + 8}{2}
    \]
    \[
    a_n = \frac{7n^2 - 13n + 8}{2}
    \]

11. To find \(a_{17}\), we substitute \(n = 17\):
    \[
    a_{17} = \frac{7(17)^2 - 13(17) + 8}{2}
    \]
    \[
    a_{17} = \frac{7 \cdot 289 - 221 + 8}{2}
    \]
    \[
    a_{17} = \frac{2023 - 221 + 8}{2}
    \]
    \[
    a_{17} = \frac{1810}{2}
    \]
    \[
    a_{17} = 905
    \]

The final answer is \(\boxed{905}\).