# 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

4n+ 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.

## 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].

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