# How to find a term in a sequence?

Arithmetic is the part of mathematics that consists of the Study of numbers and operations performed on that numbers. such operations are addition, subtraction, multiplication, division, exponentiation, and extraction of roots. The sequence is a collection/enumerated collection of objects/numbers which follow a definite pattern. Similar to a Set it can contain an infinite (possibly) number of elements (members of a sequence are called elements), and it also allowed the repetition of numbers. For example: 1, 2, 3, 4, 5, 6. The total number of elements in a sequence is called the length of a sequence. In the above example length of the sequence is 6. If there is a sequence present such as a sequence of natural numbers (1,2,3,4,5,6…), then the length of this sequence becomes infinite.

### Arithmetic Sequence

An Arithmetic Sequence is also a sequence with a definite pattern. If you take any number in the sequence then subtract it by the previous one, and the result is always the same or constant then it is an arithmetic sequence. In AP, the common difference is the constant difference in all pairs of consecutive or successive numbers in a sequence. It is represented by d. For example: 2,4,6,8. in this sequence, the common difference is d = 4-2, which is equals to 2. Similarly, 6-4 = 2, 8-6 = 2. So common difference of this sequence is 2 [Common difference= a_{2 }– a_{1 }].

Generally, there are two types of sequences present, 1. Increasing Sequence 2. Decreasing Sequence

**Increasing Sequence:-**If the common difference (d) between consecutive terms in a sequence is positive, we can say that the sequence is increasing. For example, 5, 10, 15, 20, 25. The common difference of the sequence is +5.**Decreasing Sequence:-**On the other hand, If the common difference (d) between consecutive terms in a sequence is negative then, we can say that sequence is decreasing. For example, 18, 14, 10, 6, 2. The common difference is -4.

### Finding a Term in a sequence

The first term in an arithmetic sequence is denoted as “a”, and then the common difference keeps adding to obtain the next term. The nth term of the A.P. is the term that occurs at the nth place. In the arithmetic sequence, finding the generalized formula for nth term means any term can be found, only the value of that term is needed to be put. For instance, n=1 will give a_{1}, n=3 will give a_{3}, and so on. The table given below represents terms in a sequence

a a a .. .. a a a |

For finding a term in a sequence there are two ways to determine,

**Recursive way/formula for determining a term (say r**^{th}term) in a sequence

a

_{r }= a_{r-1 }+ d for r>=2(for last (n

^{th}) term) ⇢ a_{n }= a_{n-1 }+ d for n>=2

here, a_{r }is the r^{th }term in a sequence. a_{r-1 }is the (r-1)^{th} term in a sequence and d is a common difference.

**Question: Find the next term 5, 8, 11, 14, 17, 20, ?**

**Solution:**

In this question first, we have to find common difference. common difference = a

_{2 }– a_{1 }= 8-5 =3.Now using formula, a

_{r }= a_{r-1 }+ d , here a_{r-1 }=20So, a

_{r }= 20 + 3 = 23.

**Explicit way/formula for determining a term (say r**^{th}term) in a sequence

a

_{r }= a_{1 }+ (r-1)d(for last (nth) term) ⇢ a

_{n }= a_{1 }+(n-1)d

Here, a_{r} is the r^{th} term in a sequence. a_{1} is the first term in a sequence and d is a common difference and r is the r^{th }element’s index.

### Sample Problems

**Question 1: Find the next term in the A.P. 10, 20, 30, 40, 50,?**

**Solution:**

Common difference = a

_{2}– a_{1}= 20-10 = 10Here a

_{1 }= 10 and r or n = 6Now using formula a

_{r}= a_{1}+ (r-1)d = 10 + (6-1)×10 = 60.

**Question 2: Find the next term in the following sequence, -7, -1, 5, 11, 17,?**

**Solution:**

Common difference(d) = a

_{2}– a_{1 }= -1-(-7) = -1+7 = 6.here

aand_{1 }= -7r or n = 6, here r is the r^{th }element’s index,

Note:n is the last index of the sequence. here index is last so we can say it n as well as r.Then by using formula , a

_{r}= a_{1 }+ (r-1)d = -7 + (6-1)×6 = -7 + 30 = 23.

**Question 3: Find the term in the middle of the following sequence 6, 11, ? , 21, 26.**

**Solution:**

Common difference(d) = a

_{2 }– a_{1}= 11-6 = 5 .Here, a

_{1 }= 6, and r = 3Then by using formula ,

a= 6 + (3-1)×5 = 6 + 10 =_{r}= a_{1 }+ (r-1)d16.

**Question 4: Find the first term in the following sequence, ?, -5, -1, 3, 7.**

**Solution**:

Common difference(d) = a

_{3 }– a_{2 }= -1 – (-5) = -1 +5 = 4Here , we know the last term in the given a

_{n }= 7. and a_{1 }is been to be determine. and n = 5 (last term)Then by using formula ,

a_{n}= a_{1 }+ (n-1)×d

7 = a_{1 }+ (5-1)*47 = a

_{1 }+ 16a

_{1 }= 7-16 =-9

**Question 5: Find the next term in the following sequence.** **-1/2, -5/6, -7/6,?**

**Solution:**

Common difference(d) = a

_{2}– a_{1 }= -5/6 – (-1/2) =-1/3Here, a

_{1 }= -1/2 , r or n = 4Then by using formula,

a-1/2 + (4-1)*-1/3 =_{r}= a_{1}+ (r-1)d =-3/2

**Question 6: Find the formula of a sequence with two given terms as following, a _{5} = 20 and a_{20} = 80.**

**Solution:**

For a

_{5}= 20

Using formula, a_{n }= a_{1}+(n-1)*dso ,for a

_{5}= a_{1 }+ (5-1)×da

_{5 }= a_{1}+ 4d20 = a

_{1}+4d ⇢ equation(i)similarly, for a

_{20 }= a_{1}+ (20-1)×da

_{20}= a_{1}+ 19d80 = a

_{1}+ 19d⇢equation(ii)Solve both equation by

Elimination method.multiply equation(i) by -1 and add it to equation(ii) to eliminate a_{1 },-20 = -a

_{1}– 4d (when equation(i) is multiply by -1)80 = a

_{1}+ 19dOn adding both

60 =15d

d = 4Putting value of d in equation(i)

20 = a

_{1}+ 4×420 = a

_{1}+ 16

a_{1 }= 4so a

_{1}= 4 and d = 4, the formula we’re looking for is,a

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

_{n}= 4 + (n-1)×4a

_{n}= 4 + 4n -4

a_{n}= 4d.