Arithmetic Series

We write $S_n$ for the sum of the first $n$ terms: $S_n=t_1+t_2+\cdots +t_n$. So $S_1=t_1$, $S_2=t_1+t_2$, and each partial sum is the previous one plus the next term: $S_n=S_{n-1}+t_n$. Reading that backwards gives a useful link between the two ideas, $t_n=S_n-S_{n-1}$: if you know the running totals, any single term is the difference between two consecutive totals. For example, if $S_5=75$ and $S_4=56$, then $t_5=75-56=19$. Keep the notation straight: $t_n$ is ONE value (a single term), while $S_n$ is a TOTAL of $n$ values. Mixing them up is the most common way to lose marks in this lesson.

As a child, Carl Friedrich Gauss was reportedly asked to add every number from 1 to 100. Instead of adding one at a time, he paired the first and last, the second and second-last, and so on: $1+100=101$, $2+99=101$, $3+98=101,\dots$ There are 50 such pairs, and every pair sums to the same total, $101$. Why must every pair match? Moving one step in from the front adds 1, and moving one step in from the back subtracts 1, so the pair total never changes.

The exact same pairing idea works for ANY arithmetic series, because stepping inward from the front adds $d$ while stepping inward from the back subtracts $d$. The cleanest way to see it, and to avoid worrying about an odd number of terms, is to write the sum twice: once forwards and once backwards. Line the two copies up and add them column by column. Every column adds to the same thing, $a+t_n$, and there are $n$ columns, so two copies of the sum equal $n(a+t_n)$. Divide by 2 to get one copy.

That gives the first sum formula, $S_n=\frac{n}{2}(a+t_n)$: the number of terms times the AVERAGE of the first and last term. It works even when $n$ is odd, since the doubling trick never needed whole pairs. If you don't know the last term but do know $a$ and $d$, substitute $t_n=a+(n-1)d$ to get a version that needs only $a$, $d$ and $n$. Setting $a=1$ and $d=1$ gives the famous special case $1+2+\cdots +n=\frac{n(n+1)}{2}$.

How To Derive The Formula For The Sum of an Arithmetic Series

Exam questions usually hide $n$. For $4+9+14+\cdots +99$, you must count the terms before summing: set $4+5(n-1)=99$, so $n-1=19$ and $n=20$. Counting "$(99-4)÷5=19$" gives the number of GAPS, not terms, and forgets the fence-post $+1$. The harder reverse question gives the total and asks for $n$: "how many terms of $5+8+11+\cdots$ add to 440?" Substituting into the second formula produces a quadratic in $n$, $3n^2+7n-880=0$. Solve it by factoring or the quadratic formula, then reject any negative or non-integer root, because $n$ counts terms. If no positive integer works, no number of terms gives exactly that sum.

• Seating and stacking: rows of seats that grow by 3, or a pile of logs with 1 fewer in each layer. Here $a$ is the first row or bottom layer, $d$ is the change per row, and $n$ is the number of rows. The word "total" signals a series.
• Growing savings or training plans: saving 20 dollars in week 1 and 5 dollars more each week after. "How much in week 12?" is a term, $t_{12}$; "how much altogether after 12 weeks?" is a sum, $S_{12}$.
• Sums of multiples: the multiples of 7 between 1 and 200 are $7,14,\dots ,196$, so $a=7$, $d=7$ and $n=28$, giving $S_{28}=14(7+196)=2842$.
• Falling objects: an object that falls 4.9 m in the first second, 14.7 m in the second and 24.5 m in the third follows $d=9.8$ m, so the total distance after $n$ seconds is an arithmetic series.