How Do You Find the Sum of the First n Terms of a Geometric Sequence?

For a geometric sequence with first term $a$, common ratio $r$, and $n$ terms, use

$$S_n=\frac{a(1-r^n)}{1-r}, \qquad r\neq 1.$$

If $r=1$, every term equals $a$, so $S_n=na$.

This is SL 1.3: Geometric sequences and series in IB Mathematics: Analysis and Approaches. It is core content for both SL and HL and may appear on Paper 1 or Paper 2.

Why the Formula Works

A geometric sequence has a constant common ratio between consecutive terms. Its first $n$ terms are

$$a,\ ar,\ ar^2,\ldots,\ ar^{n-1}.$$

The final exponent is $n-1$ because the first term, $a$, has exponent zero. Therefore, the sum is

$$S_n=a+ar+ar^2+\cdots+ar^{n-1}.$$

Multiply every term by $r$:

$$rS_n=ar+ar^2+\cdots+ar^{n-1}+ar^n.$$

Subtract the second equation from the first. All the middle terms cancel, leaving

$$S_n-rS_n=a-ar^n.$$

Factor both sides:

$$S_n(1-r)=a(1-r^n).$$

Dividing by $1-r$ gives the formula. The equivalent form $S_n=\frac{a(r^n-1)}{r-1}$ is equally valid and may be more convenient when .

This cancellation is the key mechanism: multiplying by $r$ shifts every term one position to the right. Subtraction then removes the repeated terms and leaves only the first term and the new final term. When identifying $r$, divide any term by the preceding term. A negative ratio is allowed and produces alternating signs; the same finite-sum formula still applies.

Worked Example

Find the sum of the first five terms of $3,6,12,24,\ldots$.

The first term is $a=3$, the ratio is $r=2$, and the number of terms is $n=5$. Thus,

$$S_5=\frac{3(1-2^5)}{1-2}=\frac{3(-31)}{-1}=93.$$

A direct check gives $3+6+12+24+48=93$.

Exam Technique

Before substituting, state $a$, $r$, and $n$ clearly; this demonstrates the method and can earn method marks even if arithmetic later goes wrong. On Paper 1, keep exact values and show algebraic working. On Paper 2, a GDC may evaluate the expression, but you should still write the formula and substitution.

A common misconception is that the $n$th term is $ar^n$; it is $ar^{n-1}$. Also, do not confuse a finite sum with $S_{\mathrm{\infty }}=\frac{a}{}$, which is valid only when .