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Geometric progressions and finite sums — 0606
Distinguish a finite sum from its convergent infinite limit.
1. Understand 2. See examples 3. Check answers 4. Explain your working Notes and key methods A geometric progression has \(u_n=ar^{n-1}\) and, for \(\displaystyle r\ne 1\), \(\displaystyle S_{n}= \frac{a(1- r^{n})}{1- r}\). For \(\displaystyle r=1\) the sum is na.
The infinite sum exists only when |r|<1 and equals \(\displaystyle \frac{a}{1-r}\). Its omitted tail after n terms is arⁿ/(1−r), not the next term alone. A ratio may be negative; alternating terms can still converge in sum.
Check the conditions The ratio is constant. An infinite sum needs |r|<1. Three worked examples Example 1 · Geometric progressions and infinite sums A geometric progression begins \(6\), \(2\). Let \(\displaystyle n=4\).
Find the ratio and a formula for the n-th term. Find \(\displaystyle S_{n}\) and explain why an infinite sum exists. Find the sum of the first n terms. Find the infinite sum. Show the reasoning (a) The ratio is the second term divided by the first; apply it n−1 times.
Answer: \(\displaystyle \frac{1}{3}\); \(6 \cdot 3^{1 - n}\)
(b) Here 0<r<1, so \(\displaystyle r^{n}\) tends to zero; use the finite formula and its limit.
Answer: \(\displaystyle \frac{80}{9}\); \(9\)
(c) Use the finite geometric-sum formula.
\(\displaystyle \frac{80}{9}\)
Answer: \(\displaystyle \frac{80}{9}\).
(d) Check |r|<1, then use \(\displaystyle \frac{a}{1-r}\).
\(9\)
Answer: \(9\).
Example 2 · Geometric progressions and infinite sums A geometric progression begins \(2\), \(\displaystyle \frac{1}{2}\). Let \(\displaystyle n=5\).
Find the ratio and a formula for the n-th term. Find \(\displaystyle S_{n}\) and explain why an infinite sum exists. Find the sum of the first n terms. Find the infinite sum. Show the reasoning (a) The ratio is the second term divided by the first; apply it n−1 times.
Answer: \(\displaystyle \frac{1}{4}\); \(2 \cdot 4^{1 - n}\)
(b) Here 0<r<1, so \(\displaystyle r^{n}\) tends to zero; use the finite formula and its limit.
Answer: \(\displaystyle \frac{341}{128}\); \(\displaystyle \frac{8}{3}\)
(c) Use the finite geometric-sum formula.
\(\displaystyle \frac{341}{128}\)
Answer: \(\displaystyle \frac{341}{128}\).
(d) Check |r|<1, then use \(\displaystyle \frac{a}{1-r}\).
\(\displaystyle \frac{8}{3}\)
Answer: \(\displaystyle \frac{8}{3}\).
Example 3 · Geometric progressions and infinite sums A geometric progression begins \(3\), \(\displaystyle \frac{3}{2}\). Let \(\displaystyle n=6\).
Find the ratio and a formula for the n-th term. Find \(\displaystyle S_{n}\) and explain why an infinite sum exists. Find the sum of the first n terms. Find the infinite sum. Show the reasoning (a) The ratio is the second term divided by the first; apply it n−1 times.
Answer: \(\displaystyle \frac{1}{2}\); \(3 \cdot 2^{1 - n}\)
(b) Here 0<r<1, so \(\displaystyle r^{n}\) tends to zero; use the finite formula and its limit.
Answer: \(\displaystyle \frac{189}{32}\); \(6\)
(c) Use the finite geometric-sum formula.
\(\displaystyle \frac{189}{32}\)
Answer: \(\displaystyle \frac{189}{32}\).
(d) Check |r|<1, then use \(\displaystyle \frac{a}{1-r}\).
\(6\)
Answer: \(6\).
Common mistakes Do not use \(\displaystyle \frac{a}{1-r}\) for every geometric progression. The tail and the next term are different quantities. Practise with answer checking Original exam-style problems Write complete solutions before opening the worked solution. The marking guidance rewards reasoning as well as answers.
Next skill: Convergent infinite geometric series →
About these materials Prepared independently from learning objectives with AI assistance for Maths4U. Mathematical checks accompany the authored models; independent teacher review is pending. Maths4U marking guidance is not an official examination-board mark scheme. Existing questions keep their original sources and results.