# Which term of the sequence 4, 9, 14, 19 is 124?

• Last Updated : 17 Aug, 2021

Sequence and Series are powerful mathematical tools used for various applications.To find any term in a sequence, first, check if the given sequence is an AP, GP, or HP, or any other series, then find the required parameters accordingly for the sequence, then find the required parameter using linear algebraic expression. Before defining the approach towards solving the given problem, first, let’s have an idea about AP or arithmetic progression.

### Arithmetic Progression

A Sequence is said to be an AP, where the differences between every two consecutive terms are the same. In progression, there is an opportunity to derive a formula for the nth term. for instance, the sequence 3, 9, 15, 21 … is a progression (AP) because it follows a pattern where each number is obtained by adding 4 to its previous term. during this sequence, nth term = 6n-3. The terms of the sequence are often obtained by substituting n=1,2,3,… within the nth term.

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• When n = 1, 6n-3 = 6(1) – 3 = 6 – 3 = 3
• When n = 2, 6n-3 = 6(2) – 3 = 12 – 3 = 9
• When n = 3, 6n-3 = 6(3) – 3 = 18 – 3 = 15

Terms used in Arithmetic Progression

• First-term, T(1)⇢ It represents the first term of the AP.
• Common Difference, (d)⇢ It represents the difference between consecutive terms of the sequence.
• Nth term of the AP⇢ This formula is used to find the nth term of the AP in the given sequence. It is defined as follows

T(n) =T(1) + (n-1)d

where n represents the nth position.

### Which term of the sequence 4, 9, 14, 19,… is 124?

The problem statement belongs to arithmetic progression, it is observed by looking at the common difference among consecutive terms, which is the same between all consecutive terms. Lets learn step by step how to solve this problem,

Analyzing the sequence

First, check if the given sequence is an arithmetic progression (AP) or not. To check for AP, calculate the common difference between the number and compare. For the given case the series is,

4, 9, 14, 19 …

Therefore,

• Common difference between 4 and 9 = |9-4| =5
• Common difference between 9 and 14 = |14-9| =5
• Common difference between 14 and 19 = |19-14| =5

As from above, it can be concluded that the common difference between numbers is the same and is equal to 5. Therefore the given sequence is an AP.

Find the parameter of an AP

The common difference of the given AP ⇢ d=5

and the first term of the  given series ⇢ T(1) = a = 4

So, in order to find an nth number of the series, use the formula

T(n) =T(1) + (n-1)×d

where T(n) denotes the nth number of the series.

Find the required parameter

In the given sequence it is asked to find the position of a given term in the sequence. Let n be the position of the term.

Given term ,T(n)= 124

Common differnec (d) =5

First term ⇢T(1) = a = 4

So by using the formula,

⇒ T(n) =T(1) + (n-1)×d

⇒ 124 =4 + (n-1)×5

⇒ 124-4 =(n-1)×5

⇒ 120 =(n-1)×5

⇒ 120/5=(n-1)

⇒ 24=(n-1)

⇒ 24+1=n

⇒ 25=n

⇒ n=25

Therefore, 124 is the 25th term in the given series.

### Similar Questions

Question 1: Given a sequence is 3, -3, -9, -15 .What is the 10th term of the sequence?

Solution:

By analyzing the sequence, see that the sequence is changing by -6 to urge to every number. The quickest way to solve this is often to stay going with the sequence by subtracting six from each number to urge to subsequent numbers. Therefore the given sequence is an Arithmetic Progression.

Common difference d = -6

First term of the series = T(1) = a = 3

So, in order to find the 10th number of the series we can use a formula:

T(n) = T(1) + (n-1)×d

For n =10

T(n) = T(1) + (n-1)×d

⇒ T(10) = T(1) + (10-1)×(-6)

⇒ T(10) = 3 + (9)×(-6)

⇒ T(10) = 3 -54

⇒ T(10) = 3 -54

⇒ T(10) = 51

Therefore the 10th term of the given sequence is 51

Question 2: The first 3 terms of a geometric sequence are 9,13 and 17. What is the 12th term?

Solution:

It’s a geometric sequence, so the difference between each consecutive term is the same.

Parameters of the sequence⇢ common difference d =13-9 = 4, first term of the series = T(1)=a= 9

So, in order to find the 12th number of the series we can use a formula:

T(n) =T(1) + (n-1)×d

For n =12

T(n) = T(1) + (n-1)×d

⇒ T(12) = T(1) + (12-1)×(4)

⇒ T(12) = 9+ (11)×(4)

⇒ T(12) = 9 + 44

⇒ T(12) = 9 + 44

⇒ T(12) = 53

Therefore the 12th term of the given sequence is 57

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