29. a) Find the formula for the sum $S_{n}$ of the first $n$ terms of an arithmetic progression.

b) Find the formula for the product $P_{n}$ of the first $n$ terms of a geometric progression.
29. a) We need to find the following sum:

$$
S_{n}=u_{1}+u_{2}+\ldots+u_{n-1}+u_{n}
$$

From problem 28 a), we know that this sum can be written as follows:

$$
\begin{aligned}
S_{n}=u_{1} & +\left(u_{1}+d\right)+\left(u_{1}+2 d\right)+\ldots+ \\
& +\left[u_{1}+(n-2) d\right]+\left[u_{1}+(n-1) d\right]
\end{aligned}
$$

This same sum can also be rewritten by rearranging the terms in reverse order:

$$
\begin{aligned}
& S_{n}=\left[u_{1}+(n-1) d\right]+\left[u_{1}+(n-2) d\right]+\ldots+ \\
&+\left(u_{1}+2 d\right)+\left(u_{1}+d\right)+u_{1} .
\end{aligned}
$$

Let's add the two obtained equations:

$$
\begin{aligned}
& 2 S_{n}=\left\{\left[u_{1}+\left[u_{1}+(n-1) d\right]\right\}+\right. \\
& \quad+\left\{\left(u_{1}+d\right)+\left[u u_{1}+(n-2) d\right]\right\}+\ldots+ \\
& \quad+\left\{\left[u_{1}+(n-2) d\right]+\left(u_{1}+d\right)\right\}+ \\
& \quad+\left\{\left[u_{1}+(n-1) d\right]+u_{1}\right\} .
\end{aligned}
$$

(Each term from the upper row is first added to the term directly below it in the lower row.) It is clear that the sum in each pair of curly braces is equal to \(2 u_{1} + (n-1) d\). We have \(n\) such pairs of curly braces. Therefore,

$$
\begin{gathered}
2 S_{n}=n\left[2 u_{1}+(n-1) d\right]=2 u_{1} n+n(n-1) d \\
S_{n}=u_{1} n+\frac{n(n-1)}{2} d .
\end{gathered}
$$

b) We need to find

$$
P_{n}=u_{1} \cdot u_{2} \cdot \ldots \cdot u_{n-1} \cdot u_{n}
$$

From problem 28 b), we know that \(P_{n}\) can be written as follows:

$$
\begin{aligned}
& P_{n}=u_{1} \cdot u_{1} q \cdot u_{1} q^{2} \cdot \ldots \cdot u_{1} q^{n-2} \cdot u_{1} q^{n-1}= \\
&=1 u_{1} q^{1} q^{1+2+\ldots}+(n-2)+(n-1)
\end{aligned}
$$

But we know (see page 11) that \(1+2+\ldots+(n-2)+(n-1)=\frac{n(n-1)}{2}\). Therefore,

$$
P_{n}=u_{1}^{n} q^{\frac{n(n-1)}{2}}
$$