Geometric progressions

In a geometric progression (G.P., or exponential sequence) you multiply by the same number every time. That number is the common ratio $r$. Find it by dividing any term by the one before: $r = T_2 \div T_1$. With first term $a$, the terms are

The first term has no $r$, the second has one, the third has $r^2$. So the $n$th term has $r$ to the power $n - 1$:

Compare an A.P., where you add $d$ each time and $T_n = a + (n - 1)d$. To tell them apart, check the differences and the ratios of the terms. For example, $2, 6, 18, 54$ has differences $4, 12, 36$, which are not equal. Its ratios are $3, 3, 3$, so it’s a G.P.

Try it

Try each ratio.

• With $r = 2$ or $3$, the terms grow faster and faster.
• With $r = \frac12$, they shrink towards 0.
• With a negative $r$, the signs alternate.

In “nth term” mode, the power on $r$ is always one less than $n$.

Finding a and r from two terms

Write each term as $ar^{n - 1}$, then divide one equation by the other. $a$ cancels and leaves a power of $r$. From $T_p$ to $T_q$ you multiply by $r$ a total of $q - p$ times, so $T_q \div T_p = r^{q - p}$.

When the power is even

If dividing leaves an even power, such as $r^2 = 4$, there are two answers: $r = 2$ or $r = -2$. Both give the same even powers. Keep both, unless the question says the ratio is positive or the terms rule one out.

Three consecutive terms

If $x$, $y$ and $z$ are consecutive terms (next to each other) of a G.P., the ratios are equal: $\dfrac{y}{x} = \dfrac{z}{y}$. Cross-multiply:

The middle term squared equals the product of its neighbours. ($y$ is the geometric mean of $x$ and $z$.)

Your turn