Just write out the cash flow.
$$\begin{align*}
AV &= 100(1+i)^{19} + 100(1+i)^{18} + \cdots + 100(1+i)^{10} \\
&\hphantom{=} + 95(1+i)^9 + 90(1+i)^8 + \cdots + 55(1+i)^1 + 50.
\end{align*}$$
Note there are $19-10+1 = 10$ payments of $100$, and the remaining $9 - 0 + 1 = 10$ payments follow the arithmetic sequence $95, 90, \ldots, 50$.
Now you can see there are multiple ways of writing out this cash flow in actuarial notation. You could do it this way:
$$\begin{align*} AV
&= 100(1+i)^{10}\left((1+i)^9 + \cdots + 1 \right) \\
&\hphantom{=} + 45(1+i)^9 + 45(1+i)^8 + \cdots + 45 \\
&\hphantom{=} + 50(1+i)^9 + 45(1+i)^8 + \cdots + 5 \\
&= 100 (1+i)^{10} s_{\overline{10}\rceil i} + 45 s_{\overline{10}\rceil i} + 5(Ds)_{\overline{10}\rceil i}
\end{align*}$$
which is the textbook solution. What they did was take the first $10$ payments as an annuity-immediate of $10$ years on $100$, and accumulated it for an additional $10$ years; then split the decreasing annuity into a level annuity-immediate of $45$, and a decreasing annuity of $5$.
But you don't have to do it this way. You can do it a number of other ways, one of which is to write
$$\begin{align*}
AV &= 100\left((1+i)^{19} + \cdots + 1\right) \\
&\hphantom{=} - \left(5(1+i)^9 + 10(1+i)^8 + \cdots + 50\right) \\
&\hphantom{=} = 100s_{\overline{20}\rceil i} - 5(Is)_{\overline{10}\rceil i}.
\end{align*}$$
In both cases, the result is the same.
I have given this advice to numerous students of actuarial science: when in doubt, write out the cash flow. It takes practice and experience to be able to directly set up the equation of value in terms of actuarial notation, and when one is still learning, skipping the cash flow step is not only more error-prone, it may actually be more time-consuming than just writing it out because you are wasting time trying to come up with how to write it compactly.
The cash flow looks like this:
$$AV = 2000\left(\left(1+\frac{i^{(4)}}{4}\right)^{\!40} \!\!\!\! + (0.98)\left(1+\frac{i^{(4)}}{4}\right)^{\!38} \!\!\!\! + (0.98)^2 \left(1+\frac{i^{(4)}}{4}\right)^{\!36} \!\!\!\!+ \cdots + (0.98)^9 \left(1 + \frac{i^{(4)}}{4}\right)^{\!22}\right)$$ where $i^{(4)} = 0.10$ is the nominal rate of interest compounded quarterly.
Explanation: the effective rate of interest per quarter period is simply $i^{(4)}/4$. To account for the payments occurring every other compounding period, we just skip those periods. Because payments are made at the beginning of each half-year, the first payment of $2000$ has had the full $10$ years, or $40$ quarters, to accumulate. To ensure that we have $5$ years of semiannual payments, or a total of $10$ payments, we require that the last payment be reduced by $(0.98)^{10 - 1}$, and that $40 - 2(9) = 22$ is the number of periods that the last payment accumulates interest.
Once you see how this is all put together, the meaning should become plainly obvious. This is why I recommend writing out the cash flow. Actuarial notation comes next. We note that we can write the above as
$$\begin{align}
AV &= 2000(1+j)^{22} \left( (1 + j)^{18} + (0.98) (1+j)^{16} + \cdots + (0.98)^9 (1+j)^0 \right) \\
&= 2000(0.98)^9 (1+j)^{22} \left( \left(\frac{(1+j)^2}{0.98}\right)^{\!9} + \left(\frac{(1+j)^2}{0.98}\right)^{\!8} + \cdots + 1 \right) \\
&= 2000(0.98)^9 (1+j)^{22} \require{enclose}s_{\enclose{actuarial}{10} j'} \\
&= 2000(0.98)^9 (1+j)^{22} \frac{(1+j')^{10} - 1}{j'},
\end{align}$$
where $j = i^{(4)}/4 = 0.025$ is the effective quarterly interest rate, and $$j' = \frac{(1+j)^2}{0.98} - 1 = \frac{113}{1568} \approx 0.072066$$ is the equivalent semiannual effective rate after adjusting for the geometric decrease in payments. It follows that $$AV \approx 40052.28.$$ The claimed answer $40042$ is inaccurate.
Alternatively, using your approach and converting the rate to a semiannual frequency, we have $j = i^{(2)}/2 = 0.050625$ as you stated, and the cash flow is then written
$$AV = 2000 \left((1 + j)^{20} + (0.98)(1 + j)^{19} + \cdots + (0.98)^9(1 + j)^{11}\right) = 2000 (0.98)^9 (1 + j)^{11} \require{enclose}s_{\enclose{actuarial}{10} j'}$$ where now $$j' = \frac{1+j}{0.98} - 1.$$ Either way gives the same result.
Best Answer
The present values of the payments can be written as $$\begin{align} PV &= a_n + a_nv + a_n v^2 + a_n v^3 +a_n v^4 + \ldots +a_n v^n\\ & = a_n(1+v+v^2+v^3+v^4+\ldots + v^n)\\ & = a_n \cdot{\ddot{a}_n} \end{align}$$
Alternatively, you could write the present value as $$ Ia_n + Da_{n-1} v^n$$ and exploit the relationships between $a_n$ and $\ddot{a}_n$ . i.e . $$\begin{align} Ia_n + Da_{n-1} v^n & = \frac{\ddot{a}_n - nv^n}{i} + \frac{n-1-a_{n-1}}{i}v^n\\ & = \frac{\ddot{a}_n-nv^n+nv^n-v^n-a_{n-1}v^n}{i}\\ & = \frac{\ddot{a}-v^n-a_{n-1}v^n}{i}\\ & = \frac{1+a_{n-1}-v^n-a_{n-1}v^n}{i}\\ & = \frac{1-v^n}{i} + \frac{a_{n-1}-a_{n-1}v^n}{i}\\ & = a_n + a_{n-1}\frac{1-v^n}{i}\\ & = a_n + a_{n-1}\cdot a_n\\ & = a_n\left(1+a_{n-1}\right)\\ & = a_n\cdot \ddot{a}_n \end{align}$$
where I have used the fact that $\ddot{a}_n = 1+ a_{n-1}$ and $a_n = \frac{1-v^n}{i}$.