Work with arithmetic progressions

Recognise arithmetic progressions

An arithmetic progression (AP) has a constant difference between consecutive terms:

$$u_{n+1}-u_n=d.$$

The common difference $d$ may be positive, negative or zero. With first term $a$, each next term is obtained by adding $d$. The source’s Models callout connects this to quantities that repeatedly change by a fixed added amount.

The unnumbered source activity has three lists: $2,7,12,17,22$ with $a=2,d=5$; $13,10,7,4,1$ with $a=13,d=-3$; and $-14,-11,-8,-5,-2$ with $a=-14,d=3$. Its follow-up question $2,-4,6,-8,10$ is not an AP: the first two consecutive differences are $-6$ and $10$. A regular-looking sign pattern is not enough to establish an AP.

The source builds the general term by matching positions to additions:

There are $n-1$ additions after the first term, so

$$u_n=a+(n-1)d,\qquad n\in\mathbb Z^+.$$

A negative difference gives a decreasing AP; a zero difference gives a constant AP. The sign of the first term does not determine the sign of the difference.

Use the arithmetic nth-term formula

Example 5 retains both complete first-term/difference/formula rows. For $-3,-8,-13,\ldots$, $a=-3,d=-5$ and $u_n=2-5n$. For $1,2\frac12,4,5\frac12,7,\ldots$, $a=1,d=3/2$ and $u_n=(3/2)n-1/2$. Self-Practice 2 then substitutes specified indices using $a=2,d=0.5$.

The next examples ask three different questions:

• Find a term (Example 6): identify $a=7,d=6$ and evaluate $u_{20}=7+19(6)=121$.
• Count terms (Example 7): treat the final value 3 as $u_n$ and solve $57+(n-1)(-3)=3$. The result is $n=19$ terms, including both endpoints. Eighteen is the number of gaps.
• Recover the progression (Example 8): translate the third and tenth terms into $a+2d=10$ and $a+9d=24$. Subtraction gives $7d=14$, hence $d=2$ and $a=6$.

In a counting problem, the resulting index must be a positive integer and really produce the stated final term. For sums of many terms, use the AP and GP sum formulas in the later methods.

Prove a progression and relate three terms

An AP test must work for every neighbouring pair in the stated domain:

$$u_{n+1}-u_n=d,$$

where the difference is independent of $n$. Equivalently, $u_n-u_{n-1}=d$ where both terms exist. This is also a recurrence relation. Checking the first two differences alone does not prove the claim for an infinite sequence.

The source challenges an argument for $u_n=2+3n$: its first terms are $5,8,11$, and their two consecutive differences are $3$. The conclusion that this is an AP is true, but that finite check is not a valid proof. Use the general neighbouring terms instead:

$$u_{n+1}-u_n=(2+3(n+1))-(2+3n)=3.$$

The general calculation completes the printed explanation. For three consecutive AP terms $x,y,z$, equating the two differences gives

$$y-x=z-y=d\quad\Longrightarrow\quad y=\frac{x+z}{2}.$$

The middle term is the arithmetic mean of its neighbours. This relation is useful when an AP problem gives information about three successive terms.