Salick Academy

Sequences and Series

45 min readNo calculatorNeeds: simultaneous equations, indices
By the end of this topic you should be able to
  • Find the nth term of an arithmetic progression
  • Find the sum of the first n terms of an arithmetic progression
  • Find the nth term and the sum of a geometric progression
  • Find the sum to infinity, and know the condition for it to exist
  • Use sigma notation

Sequences and the nth term

A sequence is an ordered list of numbers. A series is what you get when you add them up. Two kinds appear on this syllabus, and telling them apart is the first thing to do in any question:

  • an arithmetic progression adds the same amount each time — the common difference dd;
  • a geometric progression multiplies by the same amount each time — the common ratio rr.

To decide which you are looking at, take differences between consecutive terms. If they are all equal it is arithmetic. If they are not, divide instead; if those are all equal it is geometric.

Arithmetic progressions

Here aa is the first term and ll is the last term. Use the second form of SnS_n when you already know the last term, and the first form otherwise.

Two facts about a progression give you two equations, and aa and dd fall out of solving them simultaneously. That is what almost every AP question is, underneath.

Geometric progressions

The two forms of SnS_n are the same expression with the top and bottom both multiplied by 1-1. Use the left one when r<1r < 1 and the right one when r>1r > 1, purely to avoid negatives.

The sum to infinity

Add up a geometric progression forever and one of two things happens. If the terms are shrinking, the total settles on a fixed number. If they are not, it grows without limit.

r<1|r| < 1 means 1<r<1-1 < r < 1: the ratio may be negative, but its size must be under 1. A progression with r=12r = -\tfrac12 has a sum to infinity; one with r=2r = -2 does not.

Sigma notation

r=1n\displaystyle\sum_{r=1}^{n} means "add up, letting rr run from 1 to nn". So

r=112(3r2)=1+4+7++34\sum_{r=1}^{12} (3r - 2) = 1 + 4 + 7 + \cdots + 34

which is an arithmetic progression with a=1a = 1, d=3d = 3 and n=12n = 12, giving S12=122[2+33]=210S_{12} = \tfrac{12}{2}\big[2 + 33\big] = 210.

Counting the terms is where marks go. From r=1r = 1 to r=12r = 12 there are 12 terms; from r=5r = 5 to r=12r = 12 there are 125+1=812 - 5 + 1 = 8.