**How Many Terms of the Arithmetic Sequence Must be Added ?**

In this section, we will learn, how we find the number of terms to be added to get the given sum.

We have to consider the given sum as S_{n}. Instead of S_{n}, we may use one of the formulas given below.

S_{n} = (n/2) [a + l] (or)

S_{n} = (n/2) [2a + (n - 1)d]

a = first term, d = common difference and n = number of terms.

**Question 1 :**

How many terms of the AP 9, 17, 25,.......... must be taken to give a sum of 636?

**Solution :**

S_{n} = 636

a = 9, d = 17 - 9 = 6

S_{n} = (n/2) [2a + (n - 1)d]

(n/2) [2(9) + (n - 1)8] = 636

(n/2) [18 + 8n - 8] = 636

(n/2) [10 + 8n] = 636

n[5 + 4n] = 636

5n + 4n^{2} = 636

4n^{2 }+ 5n - 636 = 0

4n^{2 }- 48n + 53n - 636 = 0

4n(n - 12) + 53(n - 12) = 0

(n - 12) (4n + 53) = 0

n = 12

By solving other factor (4n + 53), we get negative value for n, which is not admissible.

Hence, 12 terms to be added to get the sum 636.

**Question 2 :**

The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and common difference.

Solution :

Given that :

First term (a) = 5, last term (l) = 45 and S_{n} = 400

S_{n} = (n/2) [a + l]

(n/2) [5 + 45] = 400

(n/2) [50] = 400

25n = 400

n = 400/25

n = 16

Hence the given series consist of 16 terms.

**Question 3 :**

The first and last term of an AP are 17 and 350 respectively.If the common difference is 9, how many terms are there and what is their sum?

**Solution :**

Given that :

First term (a) = 17, last term (l) = 350 and common difference (d) = 9 .

S_{n} = (n/2)[a + l]

By applying the values of a and l, we get

S_{n} = (n/2)[17 + 350]

S_{n} = (n/2)(367) ----(1)

S_{n} = (n/2)[2a + (n - 1)d]

S_{n} = (n/2)[2(17) + (n - 1)(9)]

S_{n} = (n/2)[34 + 9n - 9]

S_{n} = (n/2)[25 + 9n] -----(2)

(1) = (2)

(n/2)(367) = (n/2)[25 + 9n]

25 + 9n = 367

9n = 367 - 25

9n = 342

Divide each side by 9, we get

n = 342/9 = 38

So, there are 38 terms in the sequence. In order to find their sum, let us apply the value of n in (1).

S_{n} = (n/2)(367)

S_{38} = (38/2)(367)

= 19(367)

S_{38} = 6973

Hence, the required sum is 6973.

After having gone through the stuff given above, we hope that the students would have understood, how many terms of the arithmetic sequence must be added.

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