Modular Series

We can see the first sequence given is all 5 modulo 7,

Then,we have

$2x \equiv 5 \equiv -2\pmod{7}$

Because $gcd(2,7) = 1$ , so we can divide 2 on the both sides to get:

$x \equiv -1 \equiv 6 \pmod{7}$

Then we know the second sequence given is obtained by just adding 1 on the first sequence, so the second sequence are numbers that are 6 modulo 7.

So the answer is $\boxed{Yes}$

Let $S_x$ be the arithmetic progression $\{5, 12, 19, 26, \cdots \}$ . Let $S_0$ denote the first term of $S_x$ , $S_1$ the second, $S_2$ the third, and so on such that:

$S_0 = 5$

$S_1 = 5 + 7$

$S_2 = 5 + 7 + 7$

It follows that $S_x = S_0 + 7x$

Now looking $2(S_x + 1) = 2(S_0 + 1) + 14x$ notice that $14x$ is definitely a multiple of $7$ and so will leave no remainder when divided by $7$ but $2(S_0 + 1) = 12 = S_1$ which leaves a remainder of $5$ when divided by $7$ .

Therefore this is $\boxed{\text{True}}$ .