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Year 12 Maths Advanced (2027) Sequences and series

Geometric Series (Summing a GP)

20 practice questions 0 video lessons Theory + worked examples
NSW · Year 12 Mathematics Advanced · Sequences and series

Geometric series add the terms of a geometric sequence, with sum \(S_n=\dfrac{a(r^{n}-1)}{r-1}\) for \(r\neq 1\).

Part of the NSW Year 12 Mathematics Advanced course, in the Sequences and series focus area of the 2024 syllabus. Work through practice questions with fully worked solutions and video lessons, or scroll down for the theory summary and worked examples.

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Theory

A geometric series is the sum of the terms of a geometric sequence. This Year 12 Mathematics Advanced topic (MAV-12-03) uses \(S_n=\dfrac{a(r^{\,n}-1)}{r-1}\) (for \(r\neq1\)) to find sums.

The sum of the first \(n\) terms of a GP is \(S_n=\dfrac{a(r^{\,n}-1)}{r-1}=\dfrac{a(1-r^{\,n})}{1-r}\), valid whenever \(r\neq1\).

The two forms are identical — use the first when \(r>1\) and the second when \(0

Terms of a GP being summedRapidly growing bars show the terms of a geometric series being added. 1 2 3 4
A geometric series adds terms that grow by a constant ratio.
\[S_n=\dfrac{a(r^{\,n}-1)}{r-1}=\dfrac{a(1-r^{\,n})}{1-r}\]
S n equals a times (r to the n minus 1) over (r minus 1)

Method

  1. Identify \(a\), \(r\) and \(n\) (find \(n\) from the last term if needed).
  2. Choose the form: \(r>1\) use \((r^n-1)/(r-1)\); \(0
  3. Substitute and evaluate.
Example 1 — Sum, \(r>1\)
Sum the first \(6\) terms of \(2,6,18,\ldots\)
Solution
\(S_6\)\(=\)\(\dfrac{2(3^6-1)}{3-1}\)
\(=\)\(728\)
Example 2 — To a last term
Find \(1+2+4+\cdots+128\).
Solution
\(n\)\(=\)\(8\ (128=2^7)\)
\(S_8\)\(=\)\(\dfrac{2^8-1}{1}=255\)
Example 3 — Sum, \(0
Sum the first \(5\) terms of \(100,50,25,\ldots\)
Solution
\(S_5\)\(=\)\(\dfrac{100\big(1-(\tfrac12)^5\big)}{1-\tfrac12}\)
\(=\)\(193.75\)
Example 4 — Apply
For \(5+15+45+\cdots\ (a=5,r=3)\): state \(r\) and find \(S_4\).
Solution
\(r\)\(=\)\(3\)
\(S_4\)\(=\)\(\dfrac{5(3^4-1)}{2}=200\)

Common pitfalls

The exponent is \(r^{\,n}\), not \(r^{\,n-1}\) — that belongs to the term formula.
This is a finite sum. For all terms of a decaying GP, use the limiting sum.
Count the terms. Powers of \(2\) up to \(128=2^7\) means \(8\) terms.

Frequently asked questions

What is the formula for the sum of a geometric series?

It is S n equals a times (r to the n minus 1) divided by (r minus 1), for r not equal to 1, where a is the first term and r the common ratio.

When do you use 1 minus r instead of r minus 1?

They give the same answer. Use a times (1 minus r to the n) over (1 minus r) when r is between 0 and 1, to keep the numbers positive.

How do you find the number of terms in a geometric series?

Use the nth term formula a n equals a times r to the power n minus 1, set it to the last term, and solve for n.

What is the difference between a geometric series and its limiting sum?

The series formula adds a fixed number of terms. The limiting sum adds infinitely many terms and only exists when the ratio is between minus 1 and 1.