An arithmetic progression is such that the eighth term is three times the third term. Show that
the sum of the first eight terms is four times the sum of the first four terms.
Let the first term of the arithmetic progression be $a,$ and the common difference be $d.$ Then the eighth term is $a + 7d,$ and the third term is $a + 2d.$
Given that the eighth term is three times the third term, we have $(a + 7d) = 3(a + 2d),$ or $4a - d = 0.$
The sum of an arithmetic progression can be found using the formula $\frac{n}{2}(2a + (n - 1)d),$ where $n$ is the number of terms. So, the sum of the first eight terms is
\begin{align*}
\frac{8}{2}(2a + 7d) &= 4(a + 2d + a + 7d) \\
&= 4(2a + 9d).
\end{align*}Similarly, the sum of the first four terms is
\begin{align*}
\frac{4}{2}(2a + 3d) &= 2(a + a + 3d) \\
&= 4(2a + 3d).
\end{align*}Since $4a - d = 0,$ we have that $2(2a + 9d) = 4(2a + 3d),$ so the sum of the first eight terms is four times the sum of the first four terms.