For $\sigma>1$, <Orthogonality of Dirichlet characters> gives
$$
\sum_{\substack{n\geq1\\n\equiv a\pmod q}}
\frac{\Lambda(n)}{n^\sigma}
=\frac1{\varphi(q)}
\sum_{\chi\bmod q}\overline{\chi(a)}
\left(-\frac{L'(\sigma,\chi)}{L(\sigma,\chi)}\right).
$$
The principal-character term is
$$
\frac1{\sigma-1}+O_q(1).
$$
Part (a) makes every nonprincipal logarithmic derivative bounded as $\sigma\to1^+$, so
$$
\sum_{\substack{n\geq1\\n\equiv a\pmod q}}
\frac{\Lambda(n)}{n^\sigma}
=\frac1{\varphi(q)(\sigma-1)}+O_q(1)
\longrightarrow\infty.
$$
If the nondecreasing <Chebyshev function in an arithmetic progression>
$$
\psi(x;q,a)=\sum_{\substack{n\leq x\\n\equiv a\pmod q}}\Lambda(n)
$$
were bounded, the <Abel summation formula> would keep the displayed Dirichlet series bounded near $\sigma=1$. Therefore
$$
\boxed{\psi(x;q,a)\longrightarrow\infty.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 150 / 4 / b / Solution

{c}
{title2=$\psi(x;q,a)$}

For $(a,q)=1$,
$$
\psi(x;q,a)
=\sum_{\substack{n\leq x\\n\equiv a\pmod q}}\Lambda(n).
$$
The <Orthogonality of Dirichlet characters> expresses it in terms of character-twisted <Von Mangoldt functions>.

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 Incoming links: Chebyshev function in an arithmetic progression