One Formula Finds the 100th Term of a Pattern Without Listing the First 99

A sequence is a list of numbers in order. It is an arithmetic progression (AP) when each term is obtained by adding the same fixed number to the previous term. That fixed number is the common difference $d$:

$$d=a_2-a_1=a_3-a_2=\dots$$

The first term is written $a$. So an AP looks like $a,a+d,a+2d,a+3d,\dots$

This part covers testing for an AP, the nth term, counting terms, and finding $a$ and $d$.

Subtract each term from the next; if every difference is the same, the sequence is an AP, and adding that difference repeatedly gives the next terms.

Worked example 1. $3,7,11,15,\dots$

$$7-3=11-7=15-11=4\text{AP with }d=4\text{next: }19,23,27$$

Worked example 2. $-5,-2,1,\dots$: $d=3$, so the next terms are $4,7,10$.

Worked example 3. $\frac{1}{2},\frac{1}{3},\frac{1}{6},\dots$

$$\frac{1}{3}-\frac{1}{2}=-\frac{1}{6}\frac{1}{6}-\frac{1}{3}=-\frac{1}{6}\text{next: }0,-\frac{1}{6},-\frac{1}{3}$$

Worked example 4. $1,4,9,16,\dots$: differences $3,5,7$ are not equal, so not an AP.

An everyday example. Suppose a taxi charges ₹30 for the first kilometre and ₹15 for each extra kilometre. The fares for $1,2,3$ km are ₹30, ₹45, ₹60 — an AP with $d=15$.

The substance. The common difference can be negative, a fraction or even zero: $5,5,5,\dots$ is an AP with $d=0$.

**Substitute the first term $a$, the common difference $d$ and the position $n$ into $a_n=a+(n-1)d$.

Worked example 1.** Find the $20$th term of $3,7,11,\dots$

$$a_{20}=3+19\times 4=79$$

Worked example 2. Find the $15$th term of $10,7,4,\dots$

$$a_{15}=10+14\times (-3)=-32$$

Worked example 3 — from the end. Find the $5$th term from the end of $3,7,11,\dots ,99$. Counting back from the last term $l$:

$$l-(n-1)d=99-4\times 4=83$$

An everyday example. **In a school auditorium, the front row has $20$ seats and each row behind has $2$ more.** Row $20$ has

$$20+19\times 2=58\text{ seats}$$

The substance. **The formula uses $n-1$, not $n$**, because the first term has had no differences added to it yet.

**Set $a_n$ equal to the last term or the given value, solve $a+(n-1)d=a_n$ for $n$, and accept it only if $n$ is a positive whole number.

Worked example 1 — number of terms.** How many terms are in $7,13,19,\dots ,205$?

$$205=7+(n-1)\times 6n-1=33n=34$$

Worked example 2 — which term. Which term of $3,8,13,\dots$ is $78$?

$$78=3+(n-1)\times 5n-1=15n=16$$

Worked example 3 — not a term. Is $100$ a term of $3,8,13,\dots$?

$$100=3+(n-1)\times 5n-1=19.4$$

**$n$ is not a whole number, so $100$ is not a term.

An everyday example. How many two-digit numbers are multiples of $7$?** They run $14,21,\dots ,98$: $98=14+(n-1)\times 7$ gives $n=13$.

The substance. **A fractional value of $n$ proves the number is not in the AP.**

**Write each given term as $a+(n-1)d$ to get two equations, subtract to find $d$, then find $a$; the arithmetic mean of two numbers is half their sum.

Worked example 1.** The $4$th term of an AP is $11$ and the $10$th term is $29$.

$$a+3d=11a+9d=296d=18d=3,a=2$$

The AP is $2,5,8,11,\dots$; check $a_{10}=2+27=29$.

Arithmetic mean. Between $a$ and $b$ it is $\frac{a+b}{2}$, so that $a$, the mean and $b$ form an AP.

$$\text{AM of }12\text{ and }30=\frac{12+30}{2}=2112,21,30$$

Worked example 2. Find $x$ if $2x+1$, $5x-2$ and $7x+1$ are in AP.

$$2(5x-2)=(2x+1)+(7x+1)10x-4=9x+2x=6$$

The terms are $13,28,43$, with $d=15$.

An everyday example. Priya increases her monthly savings by a fixed amount. She saves ₹150 in the $3$rd month and ₹270 in the $7$th: $a+2d=150$ and $a+6d=270$ give $d=30$ and $a=90$, so she began with ₹90.

The substance. The middle one of three terms in an AP is always the arithmetic mean of the other two.