Can we solve this?

Let $a$ be the first term, $d$ be the common difference, $a_3$ be the third term and $a_7$ be the seventh term.
$a_3 = a + (3-1)d\\ a + 2d = 11 \longrightarrow \boxed{1}\\ \\ a_7 = a + (7-1)d\\ a + 6d = 27 \longrightarrow \boxed{2}$
Subtracting $\boxed{1}$ from $\boxed{2}$ , we get:-
$4d = 16\\ d = 4$
Substituting $d = 4$ in $\boxed{1}$ , we get:-
$a + 2×4 = 11\\ a = 11 - 8\\ a = 3$

So, the AP is $3,7,11,15,19,23,27,31,35,39,43,47,51,55, \cdots$ .

Here we observe that the fourth, ninth, fourteenth, ..... terms are divisible by $5$ .
They form an AP $15, 35, 55, 75, \cdots$ starting with $15$ and common difference between its terms as $20$ .
Let $n$ terms of this AP sum up to $2030$ .
Then $2030 = \dfrac{n}{2} × \left(2×15 + (n - 1)×20\right)\\ 4060 = n\left(30 + 20n - 20\right)\\ 4060 = n\left(20n - 10\right)\\ 4060 = 20n^2 + 10n\\ 20n^2 + 10n - 4060 = 0\\ 10\left(2n^2 + n - 406\right) = 0\\ 2n^2 + n - 406 = 0\\ 2n^2 + 29n - 28n - 406 = 0\\ n\left(2n + 29\right) - 14\left(2n + 29\right) = 0\\ \left(2n + 29\right)\left(n - 14\right) = 0$
So, $2n + 29 = 0$ or $n - 14 = 0$
$n = -14.5$ or $n = 14$ .
Since number of terms cannot be negative, $n \neq 14.5$ . So, $n = 14$ . So, $14$ terms of the AP $15,35,55,\cdots$ are needed to sum upto $2030$ . So, how many terms of the original AP is needed?

Observe that the first term of the 5-multiples AP is the fourth term of the original AP. The second term of the 5-multiples AP is the ninth term of the original AP. The third one is the fourteenth one. So, the $x^{\text{th}}$ term of the 5-multiples AP is the ${(5x - 1)}^{\text{th}}$ term of the original AP. So, the ${14}^{\text{th}}$ term of the 5-multiples AP is $5×14 - 1 = {69}^{\text{th}}$ term of the original AP. So, the minimum number of terms of the original AP required to satisfy the conditions of the question is $\color{#69047E}{\boxed{69}}$ .