# How to find the nth term of an Arithmetic Sequence?

Arithmetic is a part of mathematics that works with different types of numbers, fractions, applied different operations on numbers like addition, multiplication, etc. The word Arithmetic comes from the Greek word **arithmos**, which means number. Also arithmetic involves exponentiation, the calculation of percentages, finding the value of number series, logarithmic functions, and square roots, etc.

### Arithmetic Progression

There is series in arithmetic called **Arithmetic Progression (AP)**, this is a sequence of numbers, where the difference between any two consecutive terms is always the same. Let’s say, a series is 2,4, 6, 8, 10, 12,.., in this series, the difference between any two consecutive numbers is 2. If 2 is added with the previous number then the next number in the series is obtained, similarly, if 2 is subtracted from the next number, the previous number is obtained.

### The formula for finding the n^{th} term

To work with this series there are some formulas available, formulas like finding the nth term in the series, formula for finding the sum of all the terms in arithmetic series. There are formulae introduced that can help in finding the value with limited given options, for instance, finding the n^{th} term from first and the last term only. Let’s say a series A consist of some element a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6},…an

A = {a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6},… a_{n}}

- Common difference between two terms (d) = (a1-a2)
- Sum of the series(S) = (n/2)[2a + (n – 1)d]
- First term = a
- 2nd term = a+d
- 3rd term = a+2d
- similarly, Nth term = a+(n-1)d

**Steps to find the n ^{th} term**

**Step 1:** At first find the first and 2nd term, that is a_{1} and a_{2}.

**Step 2:** Then find the common difference between them, that is d = (a_{2}-a_{1})

**Step 3**: Now, by adding the difference d with the 2nd term we will get 3rd term, and like this, the series goes on. That is 2nd term, **a _{2} = a_{1}+d (a_{1} is first term)**

3rd term, **a _{3} = a_{2}+d = (a_{1}+d)+d = a_{1}+2d**

4th term, **a _{4} = a_{3}+d = (a_{1}+2d)+d = a_{1}+3d**

So, we can see that **the number of d is 1 less than the number of terms. **That is,

In a_{2}, the number of d is 1, [(2-1) = 1]

In a_{3} the number of d is 2, [(3-1) = 2]

In a_{4} the number of d is 3, [(4-1) = 3]

So, similarly, for the Nth term, the number of d must be (N-1) times.

**Therefore**, **Nth term, an = a _{1} + (N-1)d ⇢ [First term + (Last term – 1)×common difference]**

### Sample Problems

**Question 1: Find the 9th term of the given series, {1, 4, 7, 10, 13, 16,….}**

**Solution:**

Here N is 9,

First-term, a

_{1}= 12nd term, a

_{2 }= 4,3rd term a

_{3}= 74th term a

_{4}= 10Now find the common difference,

d = a

_{2}– a_{1 }= 4 – 1 = 3check d is correct or not,

a

_{1}+ d = 1+3 = 4 = a_{2}a

_{2}+ d = 4+3 = 7 = a_{3}a

_{3}+ d = 7+3 = 10 = a_{4}a

_{4}+ d = 10+3 = 13 = a_{5}a

_{5}+ d = 13+3 = 16 = a_{6}So, here the common difference is correct.

now the 9th term,

a_{9}= First term + (Last term – 1) × common difference= a1 + (N-1)d

= 1 + (9-1)×3

= 1+ 8*3

= 1+24

=25

So, the 9th term is 25.

**Question 2: Find the 12th term of the given series, {5, 11, 17, 23, 29,….}**

**Solution:**

Here N is 12,

First-term, a

_{1}= 52nd term, a

_{2}= 11,3rd term a

_{3}= 174th term a

_{4 }= 23Now find the common difference,

d = a

_{2}– a_{1}= 11 – 5 = 6check d is correct or not,

a

_{1 }+ d = 5+6 = 11 = a_{2}a

_{2}+ d = 11+6 = 17 = a_{3}a

_{3 }+ d = 17+6 = 23 = a_{4}a

_{4}+ d = 23+6 = 29 = a_{5}So, here the common difference is correct.

Now the 12th term,

a_{12}= First term + (12th term – 1) × common difference= a

_{1}+ (N-1)d= 1 + (12-1)×6

= 1+ 11×6

= 1+66

= 67

So, the 12th term is 67.

**Question 3: If the 4th term of an AP is 8 with a common difference of 2. Find out the arithmetic progression up to 8 terms.**

**Solution:**

Given that, the fourth term, a

_{4}is 8 and the common difference is 2,So the fourth term can be written as,

a + (4 – 1) × 2 = 8 [a = first term]

= a+ 3×2 = 8

= a = 8 – 3×2

= a = 8 – 6

= a = 2

So the first term a

_{1}is 2,Now, a

_{2}= a_{1}+2 = 2+2 = 4a

_{3}= a_{2}+2 = 4+2 = 6a

_{4}= 8a

_{5}= a_{4 }+2 = 8+2 = 10a

_{6}= a_{5}+2 = 10+2 = 12a

_{7}= a_{6}+2 = 12+2 = 14a

_{8}= a_{7}+2 = 14+2 = 16

So the series upto 8 term is = 2, 4, 6, 8, 10, 12, 14, 16

**Question 4: Find the 7th term of an AP whose 3rd term is 9 and 5th term is 15?**

**Solution:**

Given that, the 3rd term, a

_{3 }is 9 and 5th term, a_{5}is 15. here we have to find common difference and first term(a_{1}).a

_{3}= a_{1}+2d = 9 [from the formula] ⇢ (1)And, a

_{5}= a_{1}+4d = 15 ⇢ (2)Solve (1) and (2),

a

_{1}+2d = 9 ⇢ (1)a

_{1}+4d = 15 ⇢ (2)Let’s apply subtraction between (1) and (2)

2d = 6, d=3

So, the common difference is 3

Now put the value of d in any one equation, here put the value of d in (1)

a

_{1}+2d = 9= a

_{1}+ 2×3 = 9 [d=3]= a

_{1}=9-6= a

_{1}=6So the first term is 3

Now to find the 7th term,

Apply the formula for finding the nth term, here n=7

a

_{7}= First term + (7th term – 1) × common differencea

_{7}= a_{1 }+ (7-1)da

_{7}=3+6×3 [d=3 and a_{1}=3]a

_{7}= 21

So the 7th term is 21.

## Please

Loginto comment...