Linear Rates and Arithmetic Sums

Linear Rates Produce Arithmetic Sequences

A rate that changes linearly has equally spaced values whenever it is sampled at equally spaced times. That pattern is useful: instead of adding every rectangle height separately, we can add the heights as an arithmetic sequence. Multiplying their sum by the common rectangle width then gives the rectangle total.

For example, if a rate increases by the same amount at each successive sample, the list of sampled rates has a constant difference between consecutive terms. This is precisely an arithmetic sequence. The same idea applies when the rate decreases at a constant amount.

In this tutorial, we use equal-width subintervals on a time interval from \(0\) to \(T\). For \(n\) subintervals, the width of each rectangle is \(\Delta t=T/n\). A left sum uses the rate at each subinterval’s left endpoint; a right sum uses the rate at each right endpoint; a midpoint sum uses the rate at each subinterval’s midpoint. Each rectangle contributes its sampled rate multiplied by \(\Delta t\).

When the rate has units such as litres per minute and time has units of minutes, each rectangle contribution has units of litres. Keep the width in the calculation: adding rates alone does not give an accumulated quantity.

Why the Arithmetic-Sequence Formula Works

Suppose \(n\) terms form an arithmetic sequence. Write their sum in forward order, then write the same sum in reverse order. Each pair of terms in corresponding positions adds to the same value, namely the first term plus the last term. Adding the two written sums therefore gives \(n\) copies of that value. The two sums are equal, so twice the sum is \(n(u_1+u_n)\).

This argument works because the terms are equally spaced: for an arithmetic sequence, terms equally far from opposite ends have the same sum. In particular, it applies to the sampled values of a linear rate on an equally spaced time grid.

Setting Up the Three Sample Lists

Let a linear rate be \(r(t)=c+mt\) on the interval from \(0\) to \(T\). Here \(c\) is the rate at time zero and \(m\) is the constant rate of change. The equal subinterval width is \(\Delta t=T/n\).

The left sample times are \(0,\Delta t,\ldots,(n-1)\Delta t\). The right sample times are \(\Delta t,2\Delta t,\ldots,n\Delta t\). The midpoint sample times are \(\Delta t/2,3\Delta t/2,\ldots,(n-1/2)\Delta t\). Substituting these times into \(r(t)\) gives the corresponding rate lists.

For the left list, the first rate is \(c\) and the last is \(c+m(n-1)\Delta t\). For the right list, the first rate is \(c+m\Delta t\) and the last is \(c+mn\Delta t\). Each list is arithmetic, so the formula above gives its sum. For midpoint samples, it is convenient to add the equally spaced sample times: their average is \(T/2\), so their sum is \(nT/2\). The rate sum is therefore \(nc+mnT/2\).

Using \(\Delta t=T/n\), these expressions can also be written as:

These formulas are a compact way to compare the three calculations for this particular setup: a linear rate on \([0,T]\) divided into \(n\) equal subintervals. The examples below also show how to calculate directly from the sample lists, which helps keep the indexing and units visible.

Worked Examples

Comparing the Totals

The examples show the practical comparison: list the sample times, evaluate the rate at those times, add the resulting arithmetic sequence, and multiply by the width. For a rising rate in the first example, the three totals were \(34\), \(40\), and \(46\) litres. For the falling rate in the second, they were \(96\), \(84\), and \(72\) litres. The direction of change affects which endpoint list begins with the larger sampled rate; the arithmetic calculation itself is unchanged.

There is also a direct relationship between the three formulas for a linear rate. From the expressions above, the midpoint total is the average of the left and right totals:

This algebraic comparison is useful even when you are calculating the sums directly. It provides a check: for the same interval, linear rate, and number of equal subintervals, the midpoint total should equal the average of the left and right totals. It does not replace checking the sample values and widths in an individual problem.

A Common Arithmetic Mistake

A frequent error is to multiply the sum of the first and last terms by the number of terms without dividing by two. For the left rates \(4,7,10,13\) in the first example, that incorrect calculation would be:

But \(68\) is twice the correct rate sum, not the rate sum itself. The correct calculation is \(4(4+13)/2=34\), so the rectangle total is \(1(34)=34\) litres. A second common error is to stop at the rate sum and forget to multiply by \(\Delta t\). That shortcut gives the correct numerical total only when the width happens to be one in the chosen time units.