The 1st, 2nd, and 10th terms of an arithmetic series are equal to 1st, 4th, 17th terms of a geometric series respectively.

Question

For the arithmetic series, the 1st number is a, the common difference is d. For the geometric series, the common ratio is r. How to figure out the value of r?

Answer 1

The $1$st, $2$nd, and $10$th terms of the arithmetic series are $a$, $a+d$, and $a+9d$; similarly, the $1$st, $4$th, and $17$th terms of the geometric series are $a$, $ar^3$, and $ar^{16}$. Thus we have $a+d = ar^3$ and $a+9d = ar^{16}$, so subtracting these equations gives $8d = a\left(r^{16}-r^3\right)$. Substituting this into the first equation now yields $$\begin{align*}&a+\frac{a}{8}\left(r^{16}-r^3\right) = ar^3 \\ &\iff 1+\frac{1}{8}\left(r^{16}-r^3\right) = r^3 \quad \text{(assuming we don't have } a = d = r = 0 \text{)} \\ &\iff r^{16}-9r^3+8 = 0 \\ &\iff (r-1)\left(r^{15}+r^{14}+\dotsb+r^3-8r^2-8r-8\right) = 0,\end{align*}$$ and so either $r = 1$ (giving $d = 0$), or (numerically) $r \approx 1.07499$.