Sequences and Series • Topic 1 of 3

Arithmetic Progression (AP) and Arithmetic Mean

A sequence is an ordered list of numbers $a_1, a_2, a_3, \dots$ written in a definite order and following some rule. Each number is a term; $a_n$ is the general term (or $n$th term) and the rule that produces it is the defining formula. A sequence is finite if it stops after a fixed number of terms and infinite if it goes on forever. When you add the terms of a sequence you get a series: from the sequence $a_1, a_2, \dots, a_n$ comes the series $a_1 + a_2 + \dots + a_n$, whose value is written $S_n$.

Reading a sequence from its $n$th-term rule. If you are told $a_n = 2n + 3$, you generate terms by substituting $n = 1, 2, 3, \dots$ — giving $5, 7, 9, \dots$. This is the single most basic skill in the chapter: a compact formula stands in for the whole infinite list, exactly the way set-builder notation stands in for a set.

An Arithmetic Progression (AP) is the simplest pattern of all: each term is obtained by adding a fixed number to the one before it. That fixed number is the common difference $d$, and the starting value is the first term $a$. So an AP looks like $a,\ a+d,\ a+2d,\ a+3d, \dots$. The defining test is simple — if $a_{n} - a_{n-1}$ is the same constant for every $n$, the sequence is an AP. The general ($n$th) term follows directly:

$$a_n = a + (n-1)d$$

To add up the first $n$ terms you use the sum formula. There are two equivalent versions — use whichever fits the information you are given:

$$S_n = \dfrac{n}{2}\left[2a + (n-1)d\right] = \dfrac{n}{2}\left(a + \ell\right)$$

Here $\ell = a_n$ is the last term. The second form is Gauss's pairing idea — pair the first and last terms, the second and second-last, and so on; each pair sums to $a + \ell$, and there are $\tfrac{n}{2}$ such pairs. A handy bridge between the two formulas is that the term and the sum are linked by $a_n = S_n - S_{n-1}$, which lets you recover any term from the sums.

The Arithmetic Mean (AM) of two numbers $a$ and $b$ is the number $A$ that sits exactly between them so that $a, A, b$ form an AP. Equating the common differences ($A - a = b - A$) gives:

$$A = \dfrac{a + b}{2}$$

More generally, to insert $k$ arithmetic means between $a$ and $b$ means to find $k$ numbers that, placed between them, make the whole list an AP of $k+2$ terms. Since $b$ is then the $(k+2)$th term, the common difference is $d = \dfrac{b-a}{k+1}$, and the $i$th inserted mean is $A_i = a + i\,d$. A neat consequence: the sum of all $k$ inserted means equals $k$ times the single AM of $a$ and $b$, i.e. $k\cdot\dfrac{a+b}{2}$.

It helps to see the AP machinery beside its multiplicative cousin, the GP, which you meet in the next topic:

FeatureArithmetic ProgressionGeometric Progression
Rule between termsadd $d$multiply by $r$
$n$th term$a_n = a + (n-1)d$$a_n = ar^{n-1}$
Sum of $n$ terms$\dfrac{n}{2}[2a+(n-1)d]$$\dfrac{a(r^n-1)}{r-1}$
Mean of $a,b$$\dfrac{a+b}{2}$$\sqrt{ab}$

Deeper Insight — why the AP is the natural home of constant change: An AP is the discrete version of a straight line. Plot the terms against their position $n$ and you get equally spaced points lying on a line of slope $d$ — that is the whole geometric meaning of "common difference". This is why APs model anything growing by a fixed amount each step: simple interest, a salary that rises by a flat increment, seats increasing row by row in an auditorium. The single most useful problem-solving trick in this topic is choosing terms symmetrically about the middle. When three numbers are in AP, write them as $a-d,\ a,\ a+d$ so the $d$'s cancel when you add; for four terms use $a-3d, a-d, a+d, a+3d$ with common difference $2d$. This symmetry turns ugly simultaneous equations into one-line solutions, and it is the reason examiners can pose "three numbers in AP" puzzles that look hard but collapse instantly once you set them up the right way.

An arithmetic progression as equally spaced points climbing by the common difference d AP: equal steps of size d a a+d a+2d a+3d a+4d d Arithmetic mean A sits exactly halfway between a and b Arithmetic Mean: A = (a + b) / 2 a A b equal gap equal gap
1
Worked Example
Find the 20th term of the AP $7, 11, 15, 19, \dots$
Solution
  1. Identify $a = 7$ and $d = 11 - 7 = 4$.
  2. Apply $a_n = a + (n-1)d$ with $n = 20$.
  3. $a_{20} = 7 + (20-1)\times 4 = 7 + 76 = 83$.

Answer: $a_{20} = 83$.

2
Worked Example
How many terms of the AP $3, 8, 13, \dots$ are needed to give a sum of $1275$?
Solution
  1. Here $a = 3$, $d = 5$, and $S_n = 1275$.
  2. $S_n = \dfrac{n}{2}\left[2a + (n-1)d\right] \Rightarrow 1275 = \dfrac{n}{2}\left[6 + 5(n-1)\right]$.
  3. $2550 = n(5n + 1) \Rightarrow 5n^2 + n - 2550 = 0$.
  4. Factorise: $5n^2 + n - 2550 = (5n + 51)(n - 50) = 0$.
  5. The positive integer solution is $n = 50$.

Answer: $n = 50$ terms.

3
Worked Example
The 4th term of an AP is $11$ and the 9th term is $26$. Find the first term and the common difference.
Solution
  1. $a_4 = a + 3d = 11$ and $a_9 = a + 8d = 26$.
  2. Subtract: $(a + 8d) - (a + 3d) = 26 - 11 \Rightarrow 5d = 15 \Rightarrow d = 3$.
  3. Back-substitute: $a + 3(3) = 11 \Rightarrow a = 2$.

Answer: $a = 2$, $d = 3$.

4
Worked Example
Find the sum of all natural numbers between $100$ and $200$ that are divisible by $4$.
Solution
  1. The relevant numbers form the AP $104, 108, 112, \dots, 196$ with $a = 104$, $d = 4$, $\ell = 196$.
  2. Number of terms: $\ell = a + (n-1)d \Rightarrow 196 = 104 + 4(n-1) \Rightarrow n - 1 = 23 \Rightarrow n = 24$.
  3. $S_n = \dfrac{n}{2}(a + \ell) = \dfrac{24}{2}(104 + 196) = 12 \times 300 = 3600$.

Answer: $3600$.

5
Worked Example
Insert three arithmetic means between $3$ and $19$.
Solution
  1. Inserting $3$ means makes a $5$-term AP: $3, \_, \_, \_, 19$, so $19$ is the 5th term.
  2. $a_5 = a + 4d \Rightarrow 19 = 3 + 4d \Rightarrow d = 4$.
  3. The means are $3 + 4 = 7$, $7 + 4 = 11$, $11 + 4 = 15$.

Answer: The three arithmetic means are $7, 11, 15$.

6
Worked Example
Three numbers are in AP and their sum is $24$. If the product of the first and the third is $63$, find the numbers.
Solution
  1. Take the symmetric form $a - d,\ a,\ a + d$.
  2. Sum: $(a - d) + a + (a + d) = 3a = 24 \Rightarrow a = 8$.
  3. Product of first and third: $(a - d)(a + d) = a^2 - d^2 = 63 \Rightarrow 64 - d^2 = 63 \Rightarrow d^2 = 1 \Rightarrow d = \pm 1$.
  4. With $a = 8$, $d = 1$ gives $7, 8, 9$ (and $d = -1$ gives the same set reversed).

Answer: The numbers are $7, 8, 9$.

7
Worked Example
Write the first four terms of the sequence whose $n$th term is $a_n = \dfrac{n}{n+1}$, and state whether it is an AP.
Solution
  1. Substitute $n = 1, 2, 3, 4$: $a_1 = \tfrac{1}{2}$, $a_2 = \tfrac{2}{3}$, $a_3 = \tfrac{3}{4}$, $a_4 = \tfrac{4}{5}$.
  2. Check differences: $a_2 - a_1 = \tfrac{2}{3} - \tfrac{1}{2} = \tfrac{1}{6}$, but $a_3 - a_2 = \tfrac{3}{4} - \tfrac{2}{3} = \tfrac{1}{12}$.
  3. The differences are not equal, so the sequence is not an AP.

Answer: Terms are $\tfrac{1}{2}, \tfrac{2}{3}, \tfrac{3}{4}, \tfrac{4}{5}$; it is not an AP.

8
Worked Example
The sum of the first $n$ terms of a sequence is $S_n = 3n^2 + 2n$. Find its $n$th term and show that the sequence is an AP.
Solution
  1. Use $a_n = S_n - S_{n-1}$.
  2. $S_{n-1} = 3(n-1)^2 + 2(n-1) = 3n^2 - 6n + 3 + 2n - 2 = 3n^2 - 4n + 1$.
  3. $a_n = (3n^2 + 2n) - (3n^2 - 4n + 1) = 6n - 1$.
  4. Since $a_n = 6n - 1$ is linear in $n$, consecutive terms differ by $6$ — a constant — so it is an AP with $d = 6$.

Answer: $a_n = 6n - 1$; the sequence is an AP with $a = 5$ and $d = 6$.

9
Worked Example
Which term of the AP $21, 18, 15, \dots$ is the first negative term?
Solution
  1. Here $a = 21$, $d = -3$, so $a_n = 21 + (n-1)(-3) = 24 - 3n$.
  2. We need $a_n < 0$: $24 - 3n < 0 \Rightarrow n > 8$.
  3. The smallest integer satisfying this is $n = 9$, giving $a_9 = 24 - 27 = -3$.

Answer: The 9th term is the first negative term ($a_9 = -3$).

10
Worked Example
Find the sum of the first $n$ odd natural numbers, and verify for $n = 5$.
Solution
  1. The odd numbers $1, 3, 5, \dots$ form an AP with $a = 1$, $d = 2$.
  2. $S_n = \dfrac{n}{2}\left[2(1) + (n-1)2\right] = \dfrac{n}{2}\left[2 + 2n - 2\right] = \dfrac{n}{2}(2n) = n^2$.
  3. Check $n = 5$: $1 + 3 + 5 + 7 + 9 = 25 = 5^2$.

Answer: The sum of the first $n$ odd numbers is $n^2$ (for $n = 5$ it is $25$).

11
Worked Example
If the $7$th term of an AP is $\dfrac{1}{9}$ and the $9$th term is $\dfrac{1}{7}$, find the $63$rd term.
Solution
  1. $a + 6d = \dfrac{1}{9}$ and $a + 8d = \dfrac{1}{7}$.
  2. Subtract: $2d = \dfrac{1}{7} - \dfrac{1}{9} = \dfrac{9 - 7}{63} = \dfrac{2}{63} \Rightarrow d = \dfrac{1}{63}$.
  3. Then $a = \dfrac{1}{9} - 6\cdot\dfrac{1}{63} = \dfrac{7}{63} - \dfrac{6}{63} = \dfrac{1}{63}$.
  4. $a_{63} = a + 62d = \dfrac{1}{63} + 62\cdot\dfrac{1}{63} = \dfrac{63}{63} = 1$.

Answer: $a_{63} = 1$.

12
Worked Example
Insert four arithmetic means between $2$ and $22$.
Solution
  1. Inserting $4$ means makes a $6$-term AP, so $22$ is the 6th term.
  2. $d = \dfrac{b - a}{k + 1} = \dfrac{22 - 2}{4 + 1} = \dfrac{20}{5} = 4$.
  3. The means are $2+4=6$, $6+4=10$, $10+4=14$, $14+4=18$.

Answer: The four arithmetic means are $6, 10, 14, 18$.

Key Points

  • A sequence is an ordered list following a rule; its $n$th term $a_n$ generates every term by substituting $n = 1, 2, 3, \dots$. Adding the terms gives a series.
  • An AP adds a fixed common difference $d$ each step; its $n$th term is $a_n = a + (n-1)d$.
  • Sum of $n$ terms: $S_n = \dfrac{n}{2}\left[2a + (n-1)d\right] = \dfrac{n}{2}(a + \ell)$, where $\ell$ is the last term.
  • Term and sum are linked by $a_n = S_n - S_{n-1}$; a sum $S_n$ quadratic in $n$ always comes from an AP.
  • The arithmetic mean of $a$ and $b$ is $A = \dfrac{a+b}{2}$.
  • To insert $k$ means between $a$ and $b$, use $d = \dfrac{b-a}{k+1}$ for the resulting $(k+2)$-term AP.
  • For three terms in AP take $a-d,\,a,\,a+d$; for four take $a-3d,\,a-d,\,a+d,\,a+3d$ — the symmetry simplifies everything.
Tap an option to check your answer0 / 4
Q1.The $n$th term of an AP is:
Explanation: $a_n=a+(n-1)d$.
Q2.The common difference of $2,5,8,\dots$ is:
Explanation: $5-2=3$.
Q3.The sum of $n$ terms of an AP is:
Explanation: Standard AP sum formula.
Q4.The arithmetic mean of $a$ and $b$ is:
Explanation: AM $=\tfrac{a+b}{2}$.