Geometric Series with a General First Term

Introduction

A geometric series is built by repeatedly multiplying each term by the same common ratio. The first term sets the scale of the series: it need not be one. Once the first term, common ratio, and number of terms are known, the partial sum can be calculated directly.

For example, the terms \(5, 10, 20, 40\) form a geometric series. The first term is \(5\), and each term is twice the previous one, so the common ratio is \(2\). The sum depends not only on that ratio, but also on the starting value and how many terms are included.

First Term, Common Ratio, and Number of Terms

Definition: A finite geometric series with first term \(a\), common ratio \(r\), and \(N\) terms is the sum of the terms \(a, ar, ar^2,\ldots, ar^{N-1}\). Here \(N\) is a positive integer. Its \(k\)-th term is \(a r^{k-1}\), for \(1\le k\le N\).

The ratio describes how consecutive terms are related: each term after the first is obtained by multiplying the preceding term by \(r\). For example, if the first term is \(7\) and the ratio is \(3\), the first five terms are \(7,21,63,189,567\). The \(N\)-term partial sum is denoted by \(S_N\).

S_N=a+ar+ar^2+\cdots+ar^{N-1} =\sum_{k=1}^{N} ar^{k-1}.

The factor \(a\) appears in every term. If \(a=1\), the terms begin \(1,r,r^2,\ldots\); for a general first term, the same pattern is scaled by \(a\). The number of terms determines the last power: the first term has power \(0\), so the \(N\)-th term has power \(N-1\).

The Finite Partial-Sum Formula

Theorem: If \(r\ne 1\), the sum of the first \(N\) terms of a geometric series is \(S_N=a(1-r^N)/(1-r)\). If \(r=1\), every term equals \(a\), so \(S_N=Na\).

The formula for \(r\ne1\) includes the starting value \(a\), the ratio \(r\), and the number of terms \(N\). The condition \(r\ne1\) matters because the denominator \(1-r\) would otherwise be zero. The case \(r=1\) is simple to calculate directly: adding \(a\) to itself \(N\) times gives \(Na\).

Why the Formula Works

To see where the finite formula comes from, write the sum once and then multiply it by \(r\). Subtracting makes the middle terms cancel. This argument works for any real \(a\) and any \(r\ne1\).

Write the sum.
\(S_N=a+ar+ar^2+\cdots+ar^{N-1}\).

Multiply by the ratio.
\(rS_N=ar+ar^2+\cdots+ar^{N-1}+ar^N\).

Subtract the second equation from the first.
All shared middle terms cancel, leaving \(S_N-rS_N=a-ar^N\), or \((1-r)S_N=a(1-r^N)\).

Divide by \(1-r\).
Since \(r\ne1\), division is valid and gives \(S_N=a(1-r^N)/(1-r)\).

This also explains why the exponent is \(N\) in the numerator: after multiplying the sum by \(r\), the unmatched final term is \(ar^N\). The original sum ends at \(ar^{N-1}\), but the subtraction produces the next power, \(r^N\).

Worked Examples

Worked Example: A Series with a Ratio Greater Than One

Find the sum of the first four terms of the geometric series with first term \(4\) and common ratio \(3\). First, write the four terms by multiplying by \(3\) each time: \(4,12,36,108\). Thus \(a=4\), \(r=3\), and \(N=4\).

S_4=\frac{4(1-3^4)}{1-3} =\frac{4(1-81)}{-2} =\frac{4(-80)}{-2} =160.

Check by adding the terms directly: \(4+12=16\), \(16+36=52\), and \(52+108=160\). The formula gives the same sum.

Worked Example: A Fractional Common Ratio

Find the sum of the first four terms of a geometric series with first term \(81\) and common ratio \(1/3\). Its terms are \(81,27,9,3\), so \(a=81\), \(r=1/3\), and \(N=4\).

S_4 =\frac{81\left(1-(1/3)^4\right)}{1-1/3} =\frac{81\left(1-1/81\right)}{2/3} =\frac{81(80/81)}{2/3} =\frac{80}{2/3} =120.

Check by direct addition: \(81+27=108\), \(108+9=117\), and \(117+3=120\). This agrees with the formula.

Worked Example: The Special Case \(r=1\)

Find the sum of the first six terms when the first term is \(5\) and the common ratio is \(1\). Multiplication by \(1\) leaves each term unchanged, so the terms are \(5,5,5,5,5,5\). The formula with denominator \(1-r\) cannot be used here, so add the six equal terms:

S_6=6\cdot5=30.

Check: \(5+5+5+5+5+5=30\). Equivalently, the special-case rule \(S_N=Na\) gives \(S_6=6\cdot5=30\).

From a Finite Sum to an Infinite Sum

A finite partial sum adds a specified number of terms. An infinite geometric series is defined through the limit of its partial sums. In the case \(0\le r<1\), the powers \(r^N\) tend to zero as \(N\) increases. Applying this to the finite formula shows how the first term affects the infinite sum.

Result: If \(0\le r<1\), the infinite geometric series with first term \(a\) and common ratio \(r\) converges, and its sum is \(a/(1-r)\). This follows by taking the limit of \(S_N=a(1-r^N)/(1-r)\) as \(N\) tends to infinity.

\lim_{N\to\infty} S_N = \lim_{N\to\infty}\frac{a(1-r^N)}{1-r} = \frac{a}{1-r}.

The denominator \(1-r\) is constant in this limit, and \(r^N\to0\) for \(0\le r<1\). For instance, with \(a=12\) and \(r=1/4\), the infinite sum is \(12/(1-1/4)=16\). The initial terms are \(12,3,3/4,\ldots\); their partial sums approach \(16\). The finite formula describes each partial sum, while the infinite formula gives its limit.

A Common Mistake: Forgetting the First Term

A frequent error is to use the formula for a series beginning with \(1\) without scaling it by the actual first term. Consider the first three terms of a series with \(a=6\) and \(r=1/2\). They are \(6,3,3/2\), so their sum is \(21/2\). If the first term is accidentally omitted from the formula, one might incorrectly calculate:

\frac{1-(1/2)^3}{1-1/2} =\frac{7/8}{1/2} =\frac{7}{4}.

That result is not the sum of the given series: it is the sum for a series whose first term is \(1\). Include \(a=6\) in the formula to get the correction:

S_3 =\frac{6(1-(1/2)^3)}{1-1/2} =6\cdot\frac{7}{4} =\frac{21}{2}.

Checking against the terms confirms the answer: \(6+3+3/2=9+3/2=21/2\). Always identify the actual first term before substituting into the formula.

Choosing and Checking a Formula

Before calculating, identify three pieces of information: the first term \(a\), the common ratio \(r\), and the number of terms \(N\). The ratio can be found by dividing a term by the preceding term when the preceding term is nonzero. Then select the correct formula, paying attention to whether \(r=1\).

Situation	Sum
\(r\ne1\), \(N\) finite terms	\(S_N=a(1-r^N)/(1-r)\)
\(r=1\), \(N\) finite terms	\(S_N=Na\)
\(0\le r<1\), infinitely many terms	\(S=a/(1-r)\)

A useful check is to calculate the first few terms and confirm that they follow the stated ratio. For a finite series, direct addition of those terms can verify a formula result, especially when the number of terms is small. For a long series, also check that the exponent in the numerator matches the number of terms: \(N\) terms give the power \(N\), not \(N-1\).

The main role of the first term is to scale the series. With the same ratio and number of terms, doubling \(a\) doubles every term and therefore doubles the partial sum. The ratio controls how terms change; the term count controls where the finite sum stops. Keeping those roles distinct makes the formula easier to interpret and apply.

Check Your Understanding

For each question, identify the first term, ratio, and number of terms where relevant before calculating.

Write the first five terms of a geometric series with first term \(3\) and common ratio \(2\).
Find the sum of the first four terms when \(a=10\) and \(r=1/2\).
Why does the finite-sum formula require a separate case when \(r=1\)?
Find the sum of the first six terms when \(a=2\) and \(r=1\).
For \(a=9\) and \(r=1/3\), find the infinite sum and state why the infinite-sum formula applies.