Answer
$\dfrac{a(1-a^{10})}{1-a}+55b$
Work Step by Step
We write the sum of the first $10$ terms of the given sequence:
$$\begin{align*}
S_{10}&=(a+b)+(a^2+2b)+(a^3+3b)+\dots+(a^{10}+10b)\\
&=(a+a^2+\dots+a^{10})+(b+2b+\dots+10b).
\end{align*}$$
The terms $a, a^2, a^3,\dots,a^{10}$ form a geometric sequence, We have:
$$\begin{align*}
S_{10}&=\dfrac{a(1-a^{10})}{1-a}+b(1+2+\dots +10)\\
&=\dfrac{a(1-a^{10})}{1-a}+\dfrac{10\cdot 11}{2}b\\
&=\dfrac{a(1-a^{10})}{1-a}+55b.
\end{align*}$$
