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= a₂ – a₁ ].

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 a₁, n=3 will give a₃, and so on. The table given below represents terms in a sequence                

a1 = Represents the First term of the sequence a2 = Represents the Second term of the sequence a3 = Represents the Third term of the sequence .. .. an-1 = Represents Second last or (n-1)th term of the sequence an = Represents Last or (n)th term of the sequence an+1 = Represents (n+1)th term of the sequence

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 = a₂ – a₁ = 8-5 = 3.

Now using formula, a_r = a_r-1 + d , here a_r-1 =20

So, a_r = 20 + 3 = 23.

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

a_r = a₁ + (r-1)d

(for last (nth) term) ⇢   a_n = a₁ +(n-1)d

Here, a_r is the r^th term in a sequence. a₁ 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 = a₂ – a₁ = 20-10 = 10

Here a₁  = 10 and r or n = 6

Now using formula   a_r = a₁ + (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) = a₂ – a₁ = -1-(-7) = -1+7 = 6.

here a₁ = -7 and r 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 = a₁ + (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) = a₂ – a₁ = 11-6 = 5 .

Here, a₁ = 6, and r = 3 

Then by using formula , a_r = a₁ + (r-1)d  = 6 + (3-1)×5 = 6 + 10 = 16.

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

Solution:

Common difference(d) = a₃ – a₂ = -1 – (-5) = -1 +5 = 4

Here , we know the last term in the given a_n = 7. and a₁ is been to be determine. and n = 5 (last term)

Then by using formula , a_n= a₁ + (n-1)×d 

7 = a₁ + (5-1)*4 

7 = a₁ + 16

a₁ = 7-16 = -9

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

Solution:

Common difference(d) = a₂ – a₁ = -5/6 – (-1/2) = -1/3

Here, a₁ = -1/2 , r or n = 4 

Then by using formula, a_r = a₁ + (r-1)d = -1/2 + (4-1)*-1/3 = -3/2

Question 6: Find the formula of a sequence with two given terms as following, a₅ = 20 and a₂₀ = 80.

Solution:

For a₅ = 20

Using formula, a_n  = a₁+(n-1)*d

so ,for a₅ = a₁ + (5-1)×d 

a₅ = a₁ + 4d   

20 = a₁ +4d ⇢  equation(i)

similarly, for a₂₀ = a₁ + (20-1)×d

a₂₀ = a₁ + 19d   

80 = a₁ + 19d ⇢  equation(ii)

Solve both equation by Elimination method. multiply equation(i) by -1 and add it to equation(ii) to                                   eliminate a₁ ,

-20 = -a₁ – 4d  (when equation(i) is multiply by -1)

80 = a₁ + 19d

On adding both 

60 =15d

d = 4

Putting value of d in equation(i)

20 = a₁ + 4×4 

20 = a₁ + 16

a₁ = 4

so a₁ = 4 and d = 4, the formula we’re looking for is,

a_n = a₁ + (n-1)d

a_n = 4 + (n-1)×4

a_n = 4 + 4n -4

a_n = 4d.