Arithmetic and Geometric Series

≈ 25 min

Two families of sequences

A sequence is arithmetic when each term is found by adding a constant first difference $d$ , and geometric when each term is found by multiplying by a constant ratio $r$ .

T_n = a + (n-1)d \qquad\text{and}\qquad T_n = a\,r^{\,n-1}

where $a = T_1$ . The sums of the first $n$ terms are

S_n = \frac{n}{2}\big[2a + (n-1)d\big] \qquad\text{and}\qquad S_n = \frac{a(r^n - 1)}{r - 1},\; r \neq 1

Sigma notation

$\sum_{k=1}^{n} T_k$ means "add the terms from $k=1$ to $k=n$ ". For example

\sum_{k=1}^{5}(3k-2) = 1 + 4 + 7 + 10 + 13 = 35

The terms have first difference $3$ , so this is an arithmetic series with $a=1$ , $d=3$ , $n=5$ : $S_5 = \tfrac{5}{2}[2(1)+4(3)] = \tfrac{5}{2}(14) = 35$ . The formula and the direct addition agree.

Worked example: a stokvel that grows

A stokvel member pays in R400 in January and increases the payment by R50 each month. The payments form an arithmetic sequence with $a = 400$ , $d = 50$ . Over one year,

S_{12} = \frac{12}{2}\big[2(400) + 11(50)\big] = 6\big[800 + 550\big] = 6(1350) = \text{R}8\,100

Infinite geometric series

If $-1 < r < 1$ the powers of $r$ shrink towards zero and the sum settles on a finite value:

S_\infty = \frac{a}{1-r}

For $24 + 12 + 6 + \dots$ we have $a = 24$ , $r = \tfrac12$ , so $S_\infty = \dfrac{24}{1 - \tfrac12} = 48$ .

During a drought a borehole pump delivers 240 L in its first hour of running, 228 L in the second hour and 216 L in the third hour. The pattern continues. How many litres does it deliver in the 10th hour?

A rumour about a load-shedding schedule spreads so that 3 learners hear it in the first minute, 6 in the second minute, 12 in the third minute, and so on. How many learners have heard it in total after 8 minutes?

A leaking JoJo tank loses 18 L on the first day, 12 L on the second day, 8 L on the third day, and the pattern continues indefinitely. What is the total volume (in litres) that will eventually be lost?