2. Geometric Sequences & Series

Video coming soon4 worked examples

Theory

A geometric sequence multiplies by a constant common ratio $r$ each step. With first term $a$ , the $n$ th term is:

u_n = ar^{\,n-1}

The sum of the first $n$ terms (for $r \neq 1$ ) is:

S_n = \frac{a(r^n - 1)}{r - 1}

If $|r| < 1$ the series converges, and the sum to infinity is:

S_\infty = \frac{a}{1 - r}, \qquad |r| < 1

The Golden Rule: find $a$ and $r$ first, and remember that a sum to infinity only exists when $|r| < 1$ .

⚠️ Common Examiner Traps

Using $S_\infty$ when $|r| \geq 1$ : the series diverges and has no sum to infinity — always check the ratio first.
Power slip: the $n$ th term is $ar^{\,n-1}$ , not $ar^{\,n}$ .
Negative ratio: if the terms alternate in sign, $r$ is negative — keep the sign throughout.

Worked examples

Example 1

A geometric sequence has first term $3$ and common ratio $2$ . Find the 6th term and the sum of the first 6 terms.

Step 1: Apply the $n$ th term formula with $a=3$ , $r=2$ :

u_6 = 3 \times 2^{\,5} = 3 \times 32 = 96

Step 2: Apply the sum formula:

S_6 = \frac{3(2^6 - 1)}{2 - 1} = \frac{3(63)}{1} = 189

Example 2

A geometric series has first term $8$ and common ratio $\tfrac{1}{2}$ . Find its sum to infinity.

Step 1: Check convergence: $|r| = \tfrac{1}{2} < 1$ , so a sum to infinity exists.

Step 2: Apply the formula:

S_\infty = \frac{8}{1 - \tfrac{1}{2}} = \frac{8}{\tfrac{1}{2}} = 16

Example 3

The sum to infinity of a geometric series is $12$ and its first term is $4$ . Find the common ratio.

Step 1: Substitute into $S_\infty = \dfrac{a}{1-r}$ :

12 = \frac{4}{1 - r}

Step 2: Rearrange for $r$ :

1 - r = \frac{4}{12} = \frac{1}{3} \implies r = \frac{2}{3}

Step 3: Since $|r| = \tfrac{2}{3} < 1$ , a sum to infinity is valid, as required.

Example 4

A geometric series has first term $5$ and common ratio $1.5$ . Find the least number of terms for which the sum exceeds $500$ .

Step 1: Since $r > 1$ , use the sum formula in the form that keeps the numbers positive:

S_n = \frac{a(r^n - 1)}{r - 1} = \frac{5(1.5^n - 1)}{0.5} = 10\left(1.5^n - 1\right)

Step 2: Set up the inequality and simplify:

10\left(1.5^n - 1\right) > 500 \implies 1.5^n - 1 > 50 \implies 1.5^n > 51

Step 3: The unknown is in the index, so take logarithms of both sides. Since $\ln 1.5 > 0$ , the inequality sign is unchanged when we divide by it:

n\ln 1.5 > \ln 51 \implies n > \frac{\ln 51}{\ln 1.5} = \frac{3.9318}{0.4055} = 9.70

Step 4: $n$ must be a whole number, so the least value is $n = 10$ .

Check: $S_9 = 10(1.5^9 - 1) = 374.4$ — not enough — while $S_{10} = 10(1.5^{10} - 1) = 566.7 > 500$ ✓