The Odd Terms

The sum of $n$ consecutive terms of an arithmetic progression is given by $S = \dfrac n2(a_a + a_z)$ , where $a_a$ and $a_z$ , respectively, are the first and last terms of the $n$ terms. Therefore,

$\begin{aligned} \sum_{k=1}^{21} a_k & = \frac {21}2 (a_a + a_z) = \frac {21}2 (a_1 + a_{21}) = 693 \\ \implies \frac {a_1+a_{21}}2 & = \frac {693}{21} = 33 \\ \implies \sum_{k=0}^{10} a_{2k+1} & = \frac {11}2 (a_a + a_z) = \frac {11}2 (a_1 + a_{21}) = 11 \times 33 = \boxed{363} \end{aligned}$

Let $a_{11}$ be the middle value

$\begin{aligned} a_{1} + a_{2} + \ldots + a_{11} + \ldots + a_{20} + a_{21} &= 693 \\ (a_{11} - 10d) + (a_{11} - 9d) + \ldots + a_{11} + \ldots + (a_{11} + 9d) + (a_{11} + 10d) &= 693 \\ 21(a_{11}) &= 693 \\ a_{11} &= 33 \end{aligned}$

$\begin{aligned} a_{1} + a_{3} + \ldots + a_{11} + \ldots + a_{19} + a_{21} &= ? \\ (a_{11} - 10d) + (a_{11} - 8d) + \ldots + a_{11} + \ldots + (a_{11} + 8d) + (a_{11} + 10d) &= ? \\ 11(a_{11}) &= ? \\ 11 (33) &= \boxed{363} \end{aligned}$