Explanation

(a)
Let $t_{n}$ denotes $n$th term of an AP whose first term and common difference are $a$ and $d$ respectively.
$\therefore\ \ t_{n} = a + (n - 1)d$
Now, $\frac{t_{1} + t_{3} + t_{5} + ... + t_{2n + 1}}{t_{2} + t_{4} + t_{6} + ... + t_{2n}}$
$= \frac{\begin{array}{r}
a + (a + 2d) + (a + 4d) \\
+ ... + (a + 2nd)
\end{array}}{\begin{array}{r}
(a + d) + (a + 3d) + (a + 5d) \\
+ ... + a + (2n - 1)d
\end{array}}$
$= \frac{\frac{n + 1}{n}\lbrack a + (a + 2nd)\rbrack}{\frac{n}{2}\left\lbrack (a + d) + \left\{ a + (2n - 1)d \right\} \right\rbrack}$
$= \frac{n + 1}{n}$