Sequences and arithmetic progressions

Imagine eight weekly contributions: first 50 currency units, then 10 more each time. What is the eighth contribution, and how much has been contributed in total? These are different questions: one term of a sequence versus the sum of its terms.

Use powers and repeated operations and functions. This is hypothetical arithmetic, with no interest, fees or withdrawals; the rule covers eight weeks.

The index is input; the term is output

A numerical sequence is a function whose inputs are positive integer indices or an initial finite set of them. We start at 1: $a_1=50$ , $a_2=60$ , $a_3=70$ . A subscript gives the position, not multiplication. Order matters; repeated values are allowed.

Our domain is $n\in\{1,2,\ldots,8\}$ . There is no term numbered 2.5: the sequence identifies separate weeks, not arbitrary real durations. Its graph therefore consists of points $(n,a_n)$ , without requiring connecting segments.

A constant difference defines the progression

An arithmetic progression has a constant difference between adjacent terms: $a_{n+1}-a_n=d$ . Here $d=10$ . A recursive definition gives the first term and the step:

a_1=50,\qquad a_{n+1}=a_n+10.

Apply the step for $n=1,\ldots,7$ . Without an initial value, “add 10” does not determine the sequence: we could start at 5 or 500. The difference may be negative or zero, giving a decreasing or constant progression. Definition and recursion, OpenStax §13.2.

Why exactly n−1?

There is one step from the first term to the second, two to the third and $n-1$ to the $n$ th. Thus the explicit formula is

a_n=a_1+(n-1)d.

It computes any allowed term directly. $a_8=50+7\cdot10=120$ . Writing $50+8\cdot10$ mistakenly adds eight steps and gives the ninth term of an extended rule. If indexing starts at 0, the formula becomes $a_n=a_0+nd$ : always check the first index.

Contributions by week

Outer pair	Sum
50 + 120	170
60 + 110	170
70 + 100	170
80 + 90	170

4 × 170 = 680

Points (1,50),…,(8,120) have integer indices; no intermediate values are specified. Uniform scales. Four outer pairs of 170 total 680 currency units: the sum of contributions, not the eighth contribution.

The points lie on $a_n=10n+40$ , but only integer indices are allowed. A few starting values do not prove a progression: 50,60,70 could be followed by 100. The constant difference must be stipulated or proved for an arbitrary allowed index.

A sum in forward and reverse order

Let $S_n=a_1+a_2+\cdots+a_n$ . Write the same sum backward: $S_n=a_n+a_{n-1}+\cdots+a_1$ . Adding corresponding positions in the two lists always gives $a_1+a_n$ , since

a_j+a_{n+1-j}=2a_1+(n-1)d.

There are $n$ such terms. Hence $2S_n=n(a_1+a_n)$ and

S_n=\frac{n(a_1+a_n)}2=\frac n2\bigl(2a_1+(n-1)d\bigr).

The derivation also works for odd $n$ : we add two full lists rather than trying to split one list into a whole number of pairs. Arithmetic sums, OpenStax §13.4.

For eight weeks, $S_8=8(50+120)/2=680$ . The diagram’s outer pairs each sum to 170: 50+120, 60+110, 70+100, 80+90. Four pairs give 680. Direct addition of all eight contributions confirms it.

Contributions and accumulation follow different rules

The third contribution is $a_3=70$ , while the total through week three is $S_3=50+60+70=180$ . Our rule gives $S_n=5n^2+45n$ : accumulation is quadratic in the index although contributions form an arithmetic progression. In general, $S_n-S_{n-1}=a_n$ , not $d$ .

Identify what is indexed and whether you need one term or a sum. Then $n-1$ counts steps, and the sum formula follows from a symmetry you can reconstruct.