Properties of an Arithmetic Progression – Definition, Types & Examples

Properties of an Arithmetic Progression: A sequence is an arrangement of numbers in which one number is marked as first, another as second, another as third, and so on. A sequence in which each term differs from its preceding term by a constant is called an arithmetic progression, written as AP. 

The difference between any two successive terms in an arithmetic progression or arithmetic sequence is always the same. The common difference, indicated by \(d\), is the difference between the consecutive terms. We will go over some of the properties of arithmetic progression that we will commonly use in solving various types of arithmetical progress problems.

Properties of an Arithmetic Progression

Property I: If a constant quantity is added to or subtracted from each term of an arithmetic progression (AP), then the resulting terms of the sequence are also in AP with the same common difference \((d)\).

Proof:

Let \(\left\{ {{a_1},{a_2},{a_3},{a_4}…………..} \right\}(i)\) be an arithmetic progression with common difference \(d\).

Again, let \(k\) be any constant quantity.

If \(k\) is added to each term of the above arithmetic progression \((i)\),

Then the resulting sequence will become:

\(a_{1}+k, a_{2}+k, a_{3}+k, a_{4}+k \ldots \ldots \ldots \ldots \ldots .\)

Let \(b_{n}=a_{n}+k, n=1,2,3,4, \ldots \ldots \ldots \ldots\)
Then the new sequence will become \(b_{1}, b_{2}, b_{3}, b_{4}, \ldots \ldots \ldots \ldots \ldots\)
We have \(b_{n+1}-b_{n}=\left(a_{n+1}+k\right)-\left(a_{n}+k\right)=a_{n+1}-a_{n}=d\) for all \(n \in N,\left[\right.\) Since, \(\left.<a_{n}\right\rangle\) is a sequence with common difference \(d]\).

As a result, after adding a constant quantity \(k\) to each term of the AP, the resulting sequence is also an arithmetic progression with common difference \(d\). 

Let us look at the description below to have a better understanding of the property.

Assume that a is the first term and \(d\) is the arithmetic progression’s common difference. The arithmetic progression therefore becomes \(a, a+d, a+2d, a+3d, a+4d\), etc. By adding a constant quantity:

If each term of the arithmetic progression \(\left\{ {a,~a + d,~a + 2d,~a + 3d,~a + 4d,~…………} \right\}\)  has a constant quantity \(k\) added to it, the result is,

\(\left\{{a + k,~a + d + k,~a + 2d + k,~a + 3d + k,~a + 4d + k,~………..} \right\}~………….~\left( i\right)\)

The first term of the above sequence \((i)\) is, \((a + k)\).

The common difference of the above sequence \((i)\) is \((a+d+k)-(a+k)=d\)

Therefore, the terms of the above sequence \((i)\) form an arithmetic progression.

As a result, adding a constant quantity to each term of an arithmetic progression produces terms that are also in an arithmetic progression with the same common difference.

Learn Sum of n-Terms of Arithmetic Progression

2. By subtracting a constant quantity:

If a constant quantity \(k\) is subtracted  from each term of the arithmetic progression \(\left\{ {a,~a + d,~a + 2d,~a + 3d,~a + 4d,~…………} \right\}\) we get,

\(\left\{{a + k,~a + d + k,~a + 2d + k,~a + 3d + k,~a + 4d + k,~………..} \right\}~………….~\left( i \right)\)

The first term of the above sequence \((ii)\) is \((a-k)\).

The common difference of the above sequence \((ii)\) is, \((a+d-k)-(a-k)=d\)

As a result, the above sequence’s terms \((ii)\) constitute an arithmetic progression.

As a result, subtracting a constant number from each term of an arithmetic progression produces terms that are also in an arithmetic progression with the same common difference. 

Property II: If each term of an arithmetic progression is multiplied or divided by a non-zero constant quantity, then the resulting sequence forms an arithmetic progression.

Proof:

Let us assume \(\left\{ {{a_1},{a_2},{a_3},{a_4}~…………..}  \right\}~………..~\left( i \right)\)  be an arithmetic progression with common difference \(d\).

Again, let \(k\) be a fixed non-zero constant quantity.

Let us obtain, \(b_{1}, b_{2}, b_{3}, b_{4}, \ldots \ldots \ldots \ldots..\) be the sequence, after multiplying each term of the given AP \((i)\) by \(k\).

\(b_{1}=a_{1} k\)

\(b_{2}=a_{2} k\)

\(b_{3}=a_{3} k\)

\(b_{4}=a_{4} k\)

…………….

…………….

\(b_{n}=a_{n} k\)

…………….

…………….

Now, \(b_{n+1}-b_{n}=\left(a_{n+1} k\right)-\left(a_{n} k\right)=\left(a_{n+1}-a_{n}\right) k=d k\) for all \(n \in N,\left[\right.\) Since, \(\left\langle {{a_n}} \right\rangle \) is a sequence with common difference \(d\)].

As a result, the new sequence obtained by multiplying each term of the A.P. by a non-zero constant amount \(k\) is also an arithmetic progression with common difference \(dk\).

Let us look at the explanation below to have a better understanding of property II.

Let us assume ‘\(a\)’ be the first term and ‘\(d\)’ be the common difference of an arithmetic progression. Then, the arithmetic progression is \(\left\{ {a,~a + d,~a + 2d,~a + 3d,~a + 4d,~…………} \right\}\)

1. On multiplying a constant quantity

If a non-zero constant quantity \(k(≠0)\) is multiplied by each term of the  arithmetic progression \(\left\{ {a,~a + d,~a + 2d,~a + 3d,~a + 4d,~…………} \right\}\) we get,

\(\left\{ {ak,~ak + dk,~ak + 2dk,~ak + 3dk,~………….}\right\}~……………~\left( {iii} \right)\)

The first term of the above sequence \((iii)\) is \(ak\).

The common difference of the above sequence \((iii)\) is, \((ak+dk)-ak=dk\)

As a result, the above sequence’s terms \((iii)\) constitute an arithmetic progression.

As a result, multiplying a non-zero constant number by each term of an arithmetic progression produces terms that are also in an arithmetic progression.

2. On dividing a constant quantity

If a non-zero constant quantity \(k(≠0)\) is divided by each term of the  arithmetic progression \(\left\{ {a,~a + d,~a + 2d,~a + 3d,~a + 4d,~…………} \right\}\) we get,

\(\left\{{\frac{a}{k},\frac{a}{k} + \frac{d}{k},\frac{a}{k} + 2\frac{d}{k},\frac{a}{k} + 3\frac{d}{k}, \ldots  \ldots  \ldots  \ldots } \right\} \ldots  \ldots  \ldots  \ldots (iv)\)

The first term of the above sequence \((iv)\) is \(\frac{a}{k}\).

The common difference of the above sequence \((iv)\) is, \(\left(\frac{a}{k}+\frac{d}{k}\right)-\frac{a}{k}=\frac{d}{k}\)

As a result, the above sequence’s terms \((iv)\) constitute an arithmetic progression.

As a result, dividing a non-zero constant number by each term of an arithmetic progression produces terms that are also in arithmetic progression. 

Property III:

In an arithmetic progression of a finite number of terms, the sum of any two terms equidistant from the beginning and the end is equal to the sum of the first and last terms.

Proof:

Let us assume ‘\(a\)’ be the first term, ‘\(d\)’ be the common difference, ‘\(l\)’ be the last term and ‘\(n\)’ be the number of terms of an AP (\(n\) is finite).

The second term from the end \(=l-d\)

The third term from the end \(=l-2 d\)

The fourth term from the end \(=l-3 d\)

The \(r^{\text {th}}\) term from the end \(=l-(r-1) d\)

Again, the \(r^{\text {th }}\) term from the beginning \(=a+(r-1) d\)

Therefore, the sum of the \(r^{\text {th}}\) terms from the beginning and the end.

\(=a+(r-1) d+l-(r-1) d\)

\(=a+r d-d+l-r d+d\)

\(=a+l\)

As a result, the total of two terms that are equidistant from the start and end is always the same as or equal to the sum of the first and final terms. 

Property IV:

Three numbers \(x, y\), and \(z\) are in arithmetic progression if and only if \(2y=x+z\).

Proof:

Let us assume that, \(x, y, z\) be in arithmetic progression.

Now, common difference \(=y-x\) and again, common difference \(=z-y\)

\(\Rightarrow y-x=z-y\)

\(\Rightarrow 2 y=x+z\)

Conversely, let \(x, y, z\) be three numbers such that \(2y=x+z\). Then we prove that \(x, y, z\) are in arithmetic progression.

We have, \(2 y=x+z\)

\(\Rightarrow y+y=x+z\)

\(\Rightarrow y-x=z-y\)

\(\Rightarrow x, y, z\) are in arithmetic progression.

Property V:

A sequence is an arithmetic progression if and only if its \(n^{\text {th}}\) term is a linear expression in \(n\),
i.e. \(a_{n}=A n+B\), where \(A, B\) are two constant quantities.

Property VI:

A sequence is an arithmetic progression if and only if the sum of its first \(n\) terms is of the form \(A n^{2}+B n\), where \(A, B\) are two constant quantities that are independent of \(n\). In this case, the common difference is \(2 A\) which is \(2\) times the coefficient of \(n^{2}\).

Property VII:

A sequence is an arithmetic progression if the terms are selected at a regular interval from an arithmetic progression.

Property VIII:

If \(x, y\), and \(z\) are three consecutive terms of an arithmetic progression, then \(2y=x+z\).

Learn About Geometric Progression

Solved Examples – Properties of an Arithmetic Progression

Q.1. Write the first five terms of AP, when \(a=23\) and \(d=-5\)
Ans: The general form of AP is \(a, a+d, a+2 d \ldots\)
When \(a=23\) and \(d=-5\), the first five terms of AP are \(23,23+(-5), 23+2(-5), 23+3(-5), 23+4(-5)\)
Hence, the first five terms of AP are \(23,18,13,8\) and \(3\).

Q.2. Find the \(50^{\text {th }}\) odd number.
Ans: Odd numbers are in AP In which, \(a=1\) and \(d=2\). So, using the formula \(a_{n}=(a+(n-1) d)\) we can find the \(50^{\text {th }}\) odd number.
By substituting in the formula, we get
\(a_{50}=1+(50-1) 2\)
\(=1+(49) 2=1+98=99\)
Thus, the \(50^{\text {th }}\) odd number is \(99\).

Q.3. Find the missing term in the following AP \(5, x, 13\)
Ans: Three numbers \(x, y, z\) are in AP if \(2 y=x+z\). Given \(x=5, z=13\) therefore \(y=\frac{x+z}{2}\).
By substituting values, we get
\(y=\frac{5+13}{2}=\frac{18}{2}=9\)
Therefore, the missing term is \(9\).

Q.4. Find the sum of first \(50\) natural numbers.
Ans: By using the formula,
\(S=\left[\left(\frac{n}{2}\right) \times(a+l)\right]\)
We can find the sum. On substitution, we get
\(S=\left[\left(\frac{50}{2}\right) \times(1+50)\right]=[(25) \times(51)]=1275\)
Therefore, the sum of the first \(50\) natural numbers is \(1275\).

Q.5. Find the sum of the first \(23\) terms of the AP \(4,-3,-10, \ldots\)
Ans: \(4,-3,-10 \ldots\) is an AP where \(a=4\) and \(d=-7\)
The \(n^{\text {th }}\) term of AP \(=[a+(n-1) d]\) and Sum \(=\frac{n}{2}[2 a+(n-1) d]\)
\(a=4, d=-7\) and \(n=23\), we get the
Sum of the first \(23\) terms \(=\frac{23}{2}[2 \times 4+(23-1)(-7)]=\frac{23}{2}[8-154]\)
\(=\frac{23}{2}[-146]=-[23 \times 73]=-1679\).

Summary

The following are the characteristics of arithmetic progressions that we studied in this article: If each term of an AP is multiplied by a constant, the resultant sequence is also an AP. If each term of an AP is multiplied by a constant, the resultant sequence is also an AP. 

If each phrase of an AP is removed by a constant, the resulting sequence is likewise an AP, If each term of an AP is divided by a constant, then the resulting sequence is also an AP etc.

Frequently Asked Questions (FAQs) – Properties of an Arithmetic Progression

Q.1. What is an example of arithmetic progression?
Ans: An arithmetic progression is a set of integers with a common difference between any two consecutive integers. Another AP’s example is \(3,6,9,12,15,18,21,..\).

Q.2. What are the four common types of sequences?
Ans: The most common four types of sequences are:
1. Arithmetic sequence 
2. Geometric sequence 
3. Harmonic sequence 
4. Fibonacci series

Q.3. What does d stand for in arithmetic progression?
Ans: In an arithmetic series, \(d\) stands for the common difference. It’s the constant that’s added to each term to get to the next.
\(d=a_{n}-a_{n-1}\)

Q.4. How to find the nth term?
Ans: We can find the \(n^{\text {th}}\) term by two ways:
Explicit formula: \(a_{n}=a_{1}+d(n-1)\) or,
Recursive formula: \(a_{n}=a_{n-1}+d\)

Q.5. What are the main characteristics of arithmetic sequences?
Ans: A constant difference exists in the arithmetic sequence \(d\).
An arithmetic sequence is a set of numbers with a constant difference between subsequent terms. Any number can begin an arithmetic series, but the difference between successive terms must always be the same.

We hope this article on the Properties of an Arithmetic Progression helps you in your preparation. Do drop in your queries in the comments section if you get stuck and we will get back to you at the earliest.