Cambridge O Level Additional Mathematics · Syllabus 4037 · Series
Geometric Progression
What is Geometric Progression?
A sequence in which each term is obtained from the previous one by multiplying by the same fixed number, called the common ratio r. Its first term is a and its nth term is a times r to the power n minus one. The defining test is that the ratio of consecutive terms is constant, which may only be computed where the preceding term is not zero.
This definition is part of the Series chapter in Cambridge O Level Additional Mathematics.
Geometric Progression in context
An arithmetic progression has a constant common difference \(d\) between consecutive terms, with \(n\)th term \(u_n=a+(n-1)d\) and sum \(S_n=\dfrac n2\{2a+(n-1)d\}\); a geometric progression has a constant common ratio \(r\), with \(u_n=ar^{n-1}\) and sum \(S_n=\dfrac{a(1-r^n)}{1-r}\) for \(r\ne1\). A repeated percentage change always produces a geometric progression, never an arithmetic one, because a fixed percentage of a changing amount is not a fixed amount. An infinite geometric sum \(S_\infty=\dfrac a{1-r}\) exists only when \(\lvert r\rvert<1\); that test must be checked before the formula is used, not after.
Questions students ask about Geometric Progression
Why do you need \(u_n=a+(n-1)d\) rather than \(a+nd\)?
Because the first term has taken no steps away from itself, so \(u_1\) must equal \(a\) exactly. Testing \(a+nd\) at \(n=1\) gives \(a+d\), the second term, one step too far; \(a+(n-1)d\) correctly gives \(a\) at \(n=1\). The same step-counting reasoning gives \(u_n=ar^{n-1}\) for a geometric progression, not \(ar^n\), since \(u_1=ar^0=a\).
When does a geometric progression have a sum to infinity?
Only when \(\lvert r\rvert<1\); the sum is then \(S_\infty=\dfrac a{1-r}\). This test must be checked before the formula is used, not after, because \(\dfrac a{1-r}\) still returns a number for values of \(r\) for which no infinite sum actually exists. The test is on \(\lvert r\rvert\), not on the sign of \(r\): a negative ratio such as \(r=-\tfrac13\) still converges, since \(\left\lvert-\tfrac13\right\rvert=\tfrac13<1\).