Arithmetic Progression: nth Term, Formulas and Examples

Arithmetic Progression: Every day, we encounter a variety of sequences that allow us to predict the next term based on past terms. In the same way, a sequence of numbers is called an Arithmetic progression if the difference between two consecutive terms is always the same. In simple terms, to calculate the next number in the series, add a fixed number to the previous number.

For instance, a person’s step distance is 3,6,9,….3,6,9,…., each subsequent step differs by three from the one before it. We can tell that the next step will be at 12 by looking at this sequence. Arithmetic Progression (AP) or arithmetic sequence is the name for this sequence type.

What is an Arithmetic Progression?

An arithmetic sequence or progression is defined as a sequence of numbers in which for every pair of consecutive terms, we get the second number by adding a fixed number to the first one. The fixed number that must be added to any term of an AP to get the next term is known as the common difference of the arithmetic progression.
The sequence in which the difference between two consecutive terms is constant is known as arithmetic progression, which is abbreviated as AP

Terms Used in Arithmetic Progression

In an AP, the main terms which are used to solve mathematical problems are:
1. First Term \(\left( a \right)\)
2. Common difference \(\left( d \right)\)
3. \({n^{th}}\) Term \(\left({{a_n}} \right)\)
4. Sum of the first n terms \(\left({{s_n}} \right)\)
5. Arithmetic Mean

Formula of AP Series

An Arithmetic Progression (A.P.) is generally represented as:
\({a_1},{a_2},{a_3},{a_4},…….\)
The different terms that are used generally in Arithmetic Progression are stated above. We shall now discuss more about them.

First Term of AP

Each number associated with series is known as term. First term of the AP is known as first number of the given series. Thus, starting, or initial number is known as first term of the AP
It is generally denoted by \({a_1}\) or \(a.\)

Example:

For series \(6,13,20,27,34,….\)
The first term is \(6.\)

Common Difference in Arithmetic Progression

An arithmetic progression is a sequence, where each term, is obtained by adding a constant number to the previous term (Except first term).
Here, constant number is called as “common difference”, and it is represented by \(d.\)

Let the first term of A.P. be \({a_1},\) then
The second term is \({a_1} + d.\)
And, the third term is \({a_1} + d + d = {a_1} + 2d\)
Fourth term is \({a_1} + 2d + d = {a_1} + 3d\) and so on.

Example:

For series \(6,13,20,27,34,…..\)
The first term is \(6.\)
Here, each term is obtained by adding \(7\) to the previous term.
Thus, common difference is \(7.\)

In general, for the A.P. series is given by \({a_1},{a_2},{a_3},{a_4},……{a_n}.\)
Hence, the common difference is given by the difference between the consecutive terms of an \({\text{A}}{\text{.P}}{\text{.}}\)
So, \(d = {a_2} – {a_1} = {a_3} – {a_2} = {a_4} – {a_3} = ….. = {a_n} – {a_{n – 1}}\)

The general formula for common difference of an A.P. is given by \(d = {a_n} – {a_{n – 1}}\)

\({n^{th}}\)Term of Arithmetic Progression

Let the arithmetic progression is \(a,a + d,a + 2d,a + 3d,….\) so on.
\({a_n} = a + \left( {n – 1} \right)d\)
The \({n^{th}}\) term or last term of an arithmetic progression is given by the formula
Where, \(a – \) First term of A.P.
\(d – \)Common difference of the A.P.
\(n – \)Number of terms in an A.P.

Sum of n Terms in Arithmetic Progression

For an A.P., the sum of the first \(n\) terms can be calculated if the first term and the total number of terms are known. The formula is:
\({S_n} = \frac{n}{2}\left[{2a + \left({n – 1} \right)d} \right]\)
Here,
\(s = \)Sum of \(n\) terms of A.P.
\(n = \)Total number of terms
\(a = \) First term
\(d = \) Common difference

Derivation:

Consider an A.P. consisting \(”n”\) terms having the sequence \(a,a + d,a + 2d,…,a + \left({n – 1} \right) \times d.\)
Sum of first \(n\) terms \( = a + \left({a + d} \right) + \left({a + 2d} \right) + ….. + \left[ {a\left({n – 1} \right) \times d} \right] – – (i)\)
Writing the terms in reverse order, we get:
\(S = \left[{a + \left({n – 1} \right) \times d} \right] + \left[{a + \left({n – 2} \right) \times d} \right] + \left[{a + \left({n – 3} \right) \times d} \right] + ….. + a – – \left({ii} \right)\)
Adding both the equations term wise, we have:
\( \Rightarrow 2S = \left[{2a + \left({n – 1} \right) \times d} \right] + \left[{2a + \left( {n – 1} \right) \times d} \right] + …. + \left[{2a + \left({n – 1} \right) \times d} \right]\left({n – terms} \right)\)
\( \Rightarrow 2S = n \times \left[{2a + \left({n – 1} \right) \times d} \right]\)
\( \Rightarrow S = \frac{n}{2}\left[{2a + \left({n – 1} \right) \times d}\right]\)

Arithmetic Progression Sum Formula for First and Last Terms

When we know the first and last term of an AP, we can calculate the sum of \(n\) terms of the AP using this formula:
\({S_n} = \frac{n}{2}\left[{First\,term + Last\,term} \right]\)

Arithmetic Mean

Let three numbers \(a,b,c\) are in AP Then the middle number \(b\) is called arithmetic mean and is given by the formula:
\(b = \frac{{a + c}}{2}\)

Arithmetic Progression Formulas List

Here is a list of formulas of arithmetic progression (AP) tabulated below. These formulas are useful to solve problems based on the arithmetic progression.

Solved Examples

Q.1. In a city, a taxi charges \(₹ 2\) for the first \({\text{km,}}\) and it cost \(₹ 1.5\) more for each subsequent \({\text{km}}{\text{.}}\) How much Keerthi needs to pay the taxi driver, if she travels \({\text{5}}\,{\text{km}}\)?

Ans: Given: the taxi charge for first is \(₹2,\) and the charge for each subsequent \({\text{km}}\) is\(₹1.5.\)
So, the taxi fare for next few kilo metres is given as
\(₹2, ₹2 +₹ 1.5, ₹2 + ₹1.5 + ₹1.5…..\) so on.
It forms arithmetic progression with common difference \(₹1.5\) and first term \(₹2.\)
The sum of \(5\) terms of the AP gives the taxi fare after \(5\,{\text{km}}.\)
By using the formula:\( {S_n} = \frac{n}{2}[2a + \left({n – 1} \right)d.\)
\( \Rightarrow {S_5} = \frac{5}{2}\left({2\left( 2 \right) + \left( {5 – 1} \right)1.5} \right)\)
\( \Rightarrow {S_5} = \frac{5}{2}\left({4 + 6} \right)\)
\( \Rightarrow {S_5} = \frac{5}{2}\left({10} \right) =₹ 25\)
So, Keerthi needs to pay \(₹25\) to the taxi driver.

Q.2. Every month Mrudula kept \(₹ 30\) in her piggy bank. Every month she used to increase the amount by \(₹ 3.\) How much money will she has on her piggy bank after \(6\) months?

Ans: The amount Mrudula kept on her piggy bank is \(₹ 30,\) and, every year she increases the amount by \(₹ 3.\)
So, first term is \(₹ 30\) and the common difference is \(₹ 3.\)
The total amount she has after \(6\) months is given by using the formula:
\({S_n} = \frac{n}{2}\left[{2a + \left({n – 1} \right)d} \right]\)
\( \Rightarrow {S_n} = \frac{6}{2}\left[{2\left({30} \right) + \left({6 – 1} \right)3} \right]\)
\( \Rightarrow {S_n} = 3\left[{60 + 15} \right]\)
\( \Rightarrow {S_n} = ₹225\)
So, Mrudula has \(₹ 225\) in her piggy bank after \(6\) months.

Q.3. The first term of an arithmetic sequence is \(4\) and the tenth term is \(67.\) What is the common difference
Ans: Let the first term be \(a\) and the common difference \(d\)
Use the formula for the \(nth\) term:\({a_n} = a + \left({n – 1}\right)d\)
The first term \( = 4\)
\( \Rightarrow a = 4 – – – \left(1 \right)\)
The tenth term \( = 67\)
\( \Rightarrow {a_{10}} = a + d\left({10 – 1} \right) = 67\)
\( \Rightarrow a = 9d = 67 – – – \left( 2 \right)\)
Substitute \(a = 4\) from \(\left( 1 \right)\) into \(\left( 2 \right)\)
\( \Rightarrow 4 + 9d = 67\)
\( \Rightarrow 9d = 63\)
\( \Rightarrow d = \frac{{63}}{9} = 7\)
The common difference is \(7.\)

Q.4. What is the thirty-second term of the arithmetic sequence \( – 12, – 7, – 2,3,…\)?
Ans: This sequence has a difference of \(5\) between each pair of numbers.
The values of \(a\) and \(d\) are:
\(a = – 12\) (the first term)
\(d = 5\) (the “common difference”)
The rule can be calculated:\({a_n} = a + d\left({n – 1} \right)\)
\( \Rightarrow {a_n} = – 12 + 5\left({n – 1} \right)\)
\( \Rightarrow {a_n} = – 12 + 5n – 5\)
\( \Rightarrow {a_n} = 5n – 17\)
So, the \(32nd\) term is:
\({a_{32}} = 5 \times 32 – 17\)
\({a_{32}} = 160 – 17\)
\({a_{32}} = 143\)

Q.5. What is the sum of the first thirty terms of the arithmetic sequence: \(50,45,40,35,…\) ?
Ans: Given: \(50,45,40,35,…\)
The values of \(a,d\) and \(n\) are:
\(a = 50\) (the first term)
\(d = -5\) (common difference)
\(n = 30\) (how many terms to add up)
Using the sum of AP formula: \({S_n} = \frac{n}{2}\left({2a + \left({n – 1} \right)d} \right)\)
we get:
\({S_{30}} = \frac{{30}}{2}\left({2 \times 50 + 29 \times \left({ – 5} \right)} \right)\)
\( \Rightarrow {S_{30}} = 15\left({100 – 45} \right)\)
\( \Rightarrow {S_{30}} = 15 \times \left({ – 45} \right)\)
\( \Rightarrow {S_{30}} = – 675\)

Q.6. Rakesh is gym trainer. He started his gym session by doing \(10\) pushups on his first day. Every day he did \(5\) pushups extra than the previous day. How many pushups had he done on the last day of his first week gym session?

Ans: The number of pushups done on his first day is \(10\) and each day, he is increasing the number of pushups by \(5.\)
The number of pushups made by Rakesh represents the arithmetic progression.
Here, first term is \(10\) and common difference is \(5.\)
The last day of the week is the \(7th\) of the given sequence, and it is found by using \({a_n} = a + \left({n – 1} \right)d.\)
\( \Rightarrow {a_7} = a + 6d = 10 + 6\left( 5 \right)\)
\(\Rightarrow {a_7} = 10 + 30 = 40\)
So, Rakesh did \(40\) pushups on the last day of his first week gym session.

Summary

In this article, we have provided Arithmetic Progression definition along with all the AP formula and solved examples. An arithmetic progression is defined as a sequence of numbers in which for every pair of consecutive terms, we get the second number by adding a fixed number to the first one. The fixed number that must be added to any term of an AP to get the next term is known as the common difference of the arithmetic progression.

Frequently Asked Questions (FAQs) – Arithmetic Progression

Let’s look at some of the frequently asked questions about Arithmetic Progression:

Q.1. How many formulas are there in arithmetic progression class \(10\)?
Ans: There are mainly two formulas associated with arithmetic progression:
1. \({n^{th}}\) term of an AP
2. Sum of \(n\) terms of an AP

Q.2. What is an Arithmetic Progression?
Ans: A sequence of numbers is called an Arithmetic progression if the difference between two consecutive terms is always the same. In simple terms, to calculate the next number in the series, add a fixed number to the previous number.

Q.3. What is the sum of all odd numbers between \(1\) to \(50.\)
Ans: The odd numbers between \(1\) to \(50\) are \(1,3,5,7,9,….,49.\)
\(a = 1,d = 2,n = 25\)
\({S_n} = \frac{n}{2}\left[{2a + \left({n – 1} \right)d} \right]\)
\( \Rightarrow {S_n} = \frac{{25}}{2}\left({1 + 49} \right) = 625\)

Q.4. What is the Arithmetic Progression Formula?
Ans: The arithmetic sequence is given by \(a,a + d,a + 2d,a + 3d,….\)
Hence, the formula to find the \(nth\) term is:
\({a_n} = a + \left({n – 1} \right) \times d.\)
Sum of \(n\) terms of the \(AP = \frac{n}{2}\left[{2a + \left({n – 1} \right) \times d} \right].\)

Q.5. What is the sum of the first \(n\) natural numbers?
Ans: With the help of AP sum formula, we can calculate the sum of the first \(n\)natural numbers.
\(S = \frac{{n\left( {n + 1} \right)}}{2}\

Q.6. What is the sum of first \(n\) even numbers?
Ans: Let the sum of first n even numbers is \({S_n}.\)
\({S_n} = 2 + 4 + 6 + 8 + 10 + …. + \left({2n} \right)\)
Solving the equation using the AP sum formula, we get:
Sum of n even numbers \( = n\left({n + 1} \right)\)