ARITHMETIC PROGRESSIONS-Notes
Maths - Notes
Introduction to Sequences
- 2,5,8,11,14,… (each term increases by 3)
- 20,15,10,5,0,… (each term decreases by 5)
Examples
These patterns prepare us to understand arithmetic progressions.
ARITHMETIC PROGRESSIONS (AP)
An arithmetic progression is a list of numbers in which each term is
obtained by adding a fixed number to the preceding term except the first
term.
This fixed number is called the common difference of the AP. Remember that
it can be positive, negative or zero.
Notation
First term = \(a\)
Common difference = \(d\)
Common Difference Formula
\[d=a_2-a_1=a_3-a_2=a_n-a_{n-1}\]
General Form of an AP
- \(a =\) first term of the AP
- \(d =\) common difference (the constant added each time)
- \(a+(n-1)d=\) nth term of the AP
Finite AP
A finite AP is a sequence of numbers in which:
- The difference between consecutive terms is constant (common difference \(d\)),
- The sequence contains only \(n\) terms, ending at a last term \(l\).
- It is represented as:\[\begin{aligned}a,\ a+d,\ a+2d,\ a+3d,\ \cdots\\\cdots,\ a+(n-2d),\ a+(n-1)d\end{aligned}\]
Where:
- \(a=\) first term
- \(d=\) common difference
- \(n=\) total number of terms
- \(l=a+(n-1)d=\) last term
Example of a Finite AP: \[4,\ 9,\ 14,\ 19,\ 25\] Here,
- \(4=\) first term
- \(5=\) common difference
- \(5=\) total number of terms
- \(l=25\) last term
Infinite Arithmetic Progressions (Infinite AP)
An infinite Arithmetic Progression is an arithmetic sequence that continues without end. Its terms follow a constant difference between successive terms.
General form:
\(a,\; a+d,\; a+2d,\; a+3d,\; \cdots\)
where \(a\) is the first term and \(d\) is the common difference.
Key Features
- No fixed number of terms:
the sequence extends indefinitely. - Constant common difference:
for all \(n\), \(a_{n}-a_{n-1}=d\). - \(n\)th-term still valid:
\(a_n = a + (n-1)d\) for any positive integer \(n\). - Sum is not finite:
an infinite AP does not have a finite sum in general (unlike some geometric series).
Examples
Example 1 — Increasing AP
Sequence: \(3,\,7,\,11,\,15,\,19,\,\dots\)
Parameters: \(a=3,\; d=4\).
Example 2 — Decreasing AP
Sequence: \(40,\,35,\,30,\,25,\,20,\,\dots\)
Parameters: \(a=40,\; d=-5\).
Example 3 — Decimal step
Sequence: \(1.5,\,1.7,\,1.9,\,2.1,\,\dots\)
Parameters: \(a=1.5,\; d=0.2\).
Nth Term & Sum
Nth-term formula (valid for both finite and infinite AP):
\(a_n = a + (n-1)d\)
Note on sums: because terms continue indefinitely, an infinite AP does not have a finite total sum in the usual arithmetic sense. Terms either grow without bound or diverge to \(-\infty\) (if \(d>0\) or \(d \lt 0 \)), so \[\displaystyle \sum_{k=1}^{\infty} a_k\] is not defined as a finite number for a non-zero \(d\).
Finite vs. Infinite AP
| Feature | Finite AP | Infinite AP |
|---|---|---|
| Number of terms | Fixed (\(n\) terms) | Unlimited (no last term) |
| Last term | Exists (\(l = a+(n-1)d\)) | Does not exist |
| Sum | Finite: \(S_n = \dfrac{n}{2}[2a+(n-1)d]\) | Not finite (generally undefined) |
| Example | \(4,\,9,\,14,\,19,\,24\) | \(3,\,7,\,11,\,15,\,\dots\) |
Example-1
For the AP \(\frac{3}{2},\; \frac{1}{2},\; -\frac{1}{2},\; -\frac{3}{2},\; \cdots\) write the
first term a and the common difference d.
Solution:
First term \[a\ = \frac{3}{2}\] Common difference \[ \begin{aligned} d&=\frac{1}{2}-\frac{3}{2}\\\\ &=\frac{1-3}{2}\\\\ &=\frac{-2}{2}\\\\ &=-1 \end{aligned} \]Example-2
Check if the list of numbers form an AP? If they form an AP, write the next two terms : \[4,\; 10,\;16,\;22,\;\cdots\]
Solution:
\[ \begin{aligned} a_2-a_1=10-4=6\\ a_3-a_2=16-10=6\\ a_4-a_3=22-16=6 \end{aligned} \] we observe that \(a_{k+1}-a_k\) is same every time, hence given list of number form as AP, with common difference \(d=6\) Hence next two number are \[ \begin{aligned} 22+6=28\\ 28+6=34 \end{aligned} \]nth Term of an AP
An Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two successive terms is constant. This constant is called the common difference and is denoted by \(d\). The first term is denoted by \(a\).
Derivation of the \(n\)th-term Formula
Let the AP be: \[a,\, a+d,\, a+2d,\, a+3d,\, \cdots\]- The 1st term is \(a\)
- The 2nd term is \(a+2d\)
- The 3rd term is \(a+2d\)
Observe the pattern: to get to the \(n\)th term we add the common difference \(d\) a total of \(n-1\) times to the first term. Therefore the \(n\)th term \(a_n\) is: \[\boxed{a_n=a+(n-1)d}\] This formula holds for every positive integer \(n\).
Example-3
Determine the AP whose 3rd term is 5 and the 7th term is 9.
Solution:
\(3^{rd}\text{ term }a_3=5\) \[ \begin{align} a_3=a+(3-1)d&=5\\ =a+2d&=5\tag{1} \end{align}\] \(7^{th}\text{ term }a_7=9\) \[ \begin{align} 7=a+(7-1)d&=9\\ a_7=a+6d&=9\tag{2} \end{align} \] Subtracting equation-(1) from equation-(2) \[ \begin{aligned} a+6d-(a+2d)&=9-5\\ a+6d-a-2d&=4\\ 4d&=4\\ d&=\frac{4}{4}\\ &=1 \end{aligned} \] \[ \begin{aligned} a_3&=a+(n-1)d\\ 5&=a+(3-1)\times 1\\ 5&=a+2\\ \Rightarrow a&=5-2\\&=3 \end{aligned} \]First term \(a\)=3 and common difference \(d\)=1 hence AP is \[3,\ 4,\ 5,\ 6,\ 7,\ 8,\ \cdots\]
Example-4
How many two-digit numbers are divisible by 3?
Solution:
List of two-digit number divisible by 3 $$12,15,18....99$$ $$\begin{aligned} a&=12\\ d&=3\\ a_{n}&=99\end{aligned}$$ $$\begin{aligned} a_{n}&=a+\left( n-1\right) d\\ 99&=12+\left( n-1\right) \times 3\\ 99-12&=3n-3\\ \Rightarrow 3n-3&=87\\ 3n&=87+3\\ 3n&=90\\ n&=\dfrac{90}{3}\\ &=30\end{aligned}$$ So, there are 30 two-digit number divisible by 3Example-5
Find the 11th term from the last term (towards the first term) of the AP : 10, 7, 4, . . ., – 62.
Solution:
AP: \(10,7,4,\cdot \cdot \cdot \cdot -62\)
First term \(a=10\)
Common Difference \(d=7-10=-3\)
Last term \(l=-62\)
Let Number of terms in AP is \(n\)
Sum of First n Terms of an AP
Definition
If an AP has first term \(a\) and common difference \(d\), the sum of its first \(n\) terms is denoted by \(S_n\) and equals the total of terms from the 1st up to the \(n\)th term.
Derivation
Consider the AP:
\(a,\; a+d,\; a+2d,\; \dots,\; a+(n-1)d\)
Write the sum in natural and reverse order:
\[\small S_n = a + (a+d) + (a+2d) + \cdots + [a+(n-1)d]\]
\[\small S_n = [a+(n-1)d] + [a+(n-2)d] + \cdots + a\]
Add term-wise:
\[2S_n = n\bigl(2a + (n-1)d\bigr)\]
Therefore:
\[\boxed{\displaystyle S_n = \frac{n}{2}\bigl[2a + (n-1)d\bigr]}\]
Alternate Formula (using last term)
If the last (nth) term is \(l\), where \[l = a + (n-1)d\] then
\[\boxed{\displaystyle S_n = \frac{n}{2}(a + l)}\]
Use this when \(l\) is given — it is typically quicker.
Example-6
If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.
Solution:
Sum of 14 terms \(S_{14}=1050\)
First term \(a=10\)
Let Common Difference is \(d\)
First term \(a=10\)
Common Difference \(d=10\)
20th term of AP is 200
Example-7
How many terms of the AP : 24, 21, 18, . . . must be taken so that their sum is 78?
Solution:
AP: \(24,21,18\cdots\)
First term \(a=24\)
Common Difference \(d=21-24=-3\)
Sum \(S_n= 78\)
Let Number of terms is \(n\)
\(n=4\) or \(13\)
Both values of \(n\) are admissible. So, the number of terms is either 4 or 13.
Remarks:
- In this case, the sum of the first 4 terms = the sum of the first 13 terms = 78.
- Two answers are possible because the sum of the terms from 5th to 13th will be zero. This is because a is positive and d is negative, so that some terms will be positive and some others negative, and will cancel out each other.
Important Points
- An arithmetic progression (AP) is a list of numbers in which each term is
obtained by adding a fixed number \(d\) to the preceding term, except the first term. The fixed number
\(d\)
is called the common difference.
The general form of an AP is \(a,\ a + d,\ a + 2d,\ a + 3d, \cdots \) - A given list of numbers \(a_1,\ a_2,\ a_3,\ . . .\) is an AP, if the differences \(a_2 – a_1,\ a_3– a_2,\ a_4– a_3,\ . . .\), give the same value, i.e., if \(a_{k + 1}– a_k\) is the same for different values of \(k\).
- In an AP with first term a and common difference d, the nth term (or the general term) is given by \(a_n = a + (n – 1) d\).
- The sum of the first \(n\) terms of an AP is given by : \[S_{n}=\dfrac{n}{2}\Bigl[ 2a+\left( n-1\right) d\Bigr]\]
- If l is the last term of the finite AP, say the nth term, then the sum of all terms of the AP is given by : \[S=\frac{n}{2}(a+l)\]
