The general term and the sum of the first $n$ terms of a geometric progression, and problems on finding $a$, $r$ and $n$.
In a geometric progression (GP) each term is the previous term multiplied by the same number, the common ratio $r$.
$$\frac{U_{k+1}}{U_k} = r \quad\text{for every } k .$$
For example $3, 6, 12, 24, \dots$ has $a = 3$ and $r = 2$; $81, 27, 9, \dots$ has $r = \dfrac{1}{3}$; $5, -10, 20, \dots$ has $r = -2$ (the signs alternate). To test for a GP, check that the ratios of consecutive terms are all equal.
Start at $a$ and multiply by $r$ each time: $U_1 = a$, $U_2 = ar$, $U_3 = ar^2$. The $n$th term has been multiplied $n - 1$ times: $$U_n = ar^{n-1} .$$
Example: Find the 8th term and the sum of the first $8$ terms of $3,\ 6,\ 12,\ \dots$. Here $a = 3$, $r = 2$, so $U_8 = 3 \times 2^7 = 384$. By the formula below, $S_8 = \dfrac{3(2^8 - 1)}{2 - 1} = 3 \times 255 = 765$.
Let $S_n = a + ar + ar^2 + \dots + ar^{n-1}$. Multiply by $r$: $$rS_n = ar + ar^2 + \dots + ar^{n-1} + ar^n .$$ Subtract. All the middle terms cancel: $S_n - rS_n = a - ar^n$, so $(1 - r)S_n = a(1 - r^n)$ and $$S_n = \frac{a(1 - r^n)}{1 - r} = \frac{a(r^n - 1)}{r - 1}, \qquad r \ne 1 .$$ Use the first form when $|r| < 1$ and the second when $r > 1$, so that the numbers stay positive. If $r = 1$ every term equals $a$ and $S_n = na$.
Example: The second term of a GP is $6$ and the fifth term is $162$. Find $a$ and $r$, and $S_6$. Dividing, $\dfrac{U_5}{U_2} = r^3 = \dfrac{162}{6} = 27$, so $r = 3$ and $a = \dfrac{6}{3} = 2$. Then $S_6 = \dfrac{2(3^6 - 1)}{3 - 1} = 3^6 - 1 = 728$.
Example: Find the least $n$ for which $1 + 2 + 4 + \dots + 2^{n-1}$ exceeds $1000$. The sum is $S_n = 2^n - 1$, so we need $2^n > 1001$. Since $2^9 = 512$ and $2^{10} = 1024$, the least value is $n = 10$ (the sum is then $1023$).
Write three numbers in GP as $\dfrac{p}{r},\ p,\ pr$; their product is $p^3$. A series such as $\displaystyle\sum_{k=1}^{n}(2^k + 3)$ is split by the properties of $\Sigma$ into a GP part and a constant part: $\sum 2^k + \sum 3 = (2^{n+1} - 2) + 3n$.
To find the ratio from two terms, divide the later by the earlier: if $U_p$ and $U_q$ ($q > p$) are known then $r^{q-p} = \dfrac{U_q}{U_p}$. Remember that this may have two solutions when $q - p$ is even (positive and negative $r$).
Exam tip: A typical question gives two terms of a GP and asks for $S_n$, or gives $S_n$ and asks for the terms. Use $\dfrac{U_q}{U_p} = r^{q-p}$ to find $r$, then $a$. Take care with a negative ratio: $(-2)^7 = -128$, so $S_7 = \dfrac{5(1 - (-2)^7)}{3}$ uses the sign carefully. Choose the form of the sum formula that avoids negative numbers where you can.