- Home/
- CTET & State TET (Earlier - TEACHING)/
- General /
- Article
How Many Terms of the AP : 9, 17, 25, . . . must be taken to Give a Sum of 636?
By BYJU'S Exam Prep
Updated on: September 25th, 2023
To find: the total term of AP: 9, 17, 25, . . .
Given: Sn = 636
Here, the first term (a) of AP is 9 and the common difference (d) is 71 – 9 = 8.
Now, we know that, Sn = n/2 [2a + (n – 1) d] or Sn = n/2 [a + 1] is the formula for calculating the sum of the first n terms of an AP.
Now, after putting the values in the formula, we will get
n/2 [2 × 9 + (n – 1) 8] = 636
n/2 [18 + 8n – 8] = 636
n/2 [10 + 8n] = 636
n[5 + 4n] = 636
5n + 4n2= 636
4n2 + 5n – 636 = 0
4n2 + 53n – 48n – 636 = 0
n (4n + 53) – 12 (4n + 53) = 0
(4n + 53)(n – 12) = 0
Either 4n + 53 = 0 or n – 12 = 0
n = – 53/4 or n = 12
Since, we have to obtain the total number of terms hence, the value of ‘n’ cannot be in fraction. So, the total number of terms for the given AP: 9, 17, 25, . . . whose sum is 636, is 12.
Table of content
Total Terms of AP 9, 17, 25, . . . whose Sum is 636
As discussed above, 12 is the total number of terms of the AP 9, 17, 25, . . . to give a sum of 636. To solve the question, it is important to understand what AP is. AP in mathematics stands for Arithmetic Progression.
- AP is a set of numbers where the difference between each term from its predecessor is fixed throughout the entire sequence.
- The common difference of that arithmetic progression is the constant difference.
- In other words, A list of numbers is an arithmetic progression if every term, with the exception of the first, is obtained by adding a fixed number to the term before it.
- This fixed number is referred to as the common difference of an AP. The common difference can be positive, negative, or zero.
- It is represented by ‘d’ in the formula of AP.
Summary:
How Many Terms of the AP 9, 17, 25, . . . must be taken to Give a Sum of 636?
The total terms of AP 9, 17, 25, . . . that must be taken to give a sum of 636 is 12. It can be determined by putting the given values in the formula to calculate the sum of first ‘n’ terms of an AP i.e. Sn = n/2 [2a + (n – 1) d] or Sn = n/2 [a + 1].