# Find the sum of the following APs:

(i) 2, 7, 12,... , to 10 terms.

(ii) - 37, - 33, - 29,....., to 12 terms.

(iii) 0.6, 1.7, 2.8,...... , to 100 terms.

(iv) 1/15 , 1/12 , 1/10 ,...., to 11 terms

**Solution:**

The sum of n terms of the AP is denoted by Sₙ. It is the total sum of all the terms of the given AP.

The sum of the first n terms of an AP is given by Sₙ = n/2 [2a + (n - 1) d]

where a is the first term, d is the common difference and n is the number of terms.

(i) 2, 7, 12,... , to 10 terms.

Given,

- First term, a = 2
- Common Difference, d = 7 - 2 = 5
- Number of Terms, n =10

We know that sum up to n^{th} term of AP is given by formula Sₙ = n/2 [2a + (n - 1) d]

S₁₀ = 10/2 [2 × 2 + (10 - 1) 5]

= 5 [4 + 9 × 5]

= 5[4 + 45]

= 5 × 49

= 245

The sum of 10 terms is 245.

(ii) - 37, - 33, - 29,......, to 12 terms.

Given,

- First term, a = - 37
- Common Difference, d = (- 33) - (- 37) = 4
- Number of Terms, n = 12

We know that sum up to n^{th} term of AP is given by formula Sₙ = n/2 [2a + (n - 1) d]

S₁₂ = 12/2 [2 (- 37) + (12 - 1) 4]

= 6 [- 74 + 11 × 4]

= 6 [- 74 + 44]

= 6 × (- 30)

= - 180

The sum of 12 terms is - 180.

(iii) 0.6, 1.7, 2.8,...... , to 100 terms.

Given,

- First term, a = 0.6
- Common Difference, d = 1.7 - 0.6 = 1.1
- Number of Terms, n = 100

We know that sum up to n^{th} term of AP is givem by formula Sₙ = n/2 [2a + (n - 1) d]

S₁₀₀ = 100/2 [2 × 0.6 + (100 - 1) 1.1]

= 50 [1.2 + 99 × 1.1]

= 50 [1.2 + 108.9]

= 50 [110.1]

= 5505

The sum of 100 terms is 5505.

(iv) 1/15 , 1/12 , 1/10 ,...., to 11 terms

Given,

- First term, a = 1/15
- Common Difference, d = 1/12 - 1/15 = 1/60
- Number of Terms, n = 11

We know that sum up to n^{th} term of AP is Sₙ = n/2 [2a + (n - 1) d]

S₁₁ = 11/2 [2 × 1/15 + (11 - 1) × 1/60]

= 11/2 [2/15 + 1/6]

= 11/2 [ (4 + 5) / 30]

= 11/2 × 3/10

= 33/20

**☛ Check: **NCERT Solutions Class 10 Maths Chapter 5

**Video Solution:**

## Find the sum of the following APs. (i) 2, 7, 12,... , to 10 terms. (ii) - 37, - 33, - 29,....., to 12 terms. (iii) 0.6, 1.7, 2.8,...... , to 100 terms. (iv) 1/15 , 1/12 , 1/10 ,...., to 11 terms

Class 10 Maths NCERT Solutions Chapter 5 Exercise 5.3 Question 1

**Summary:**

The sum of the following A.P's is as follows: For the first A.P., 2, 7, 12,... , to 10 terms, the sum is 245, for the second A.P., - 37, - 33, - 29,....., to 12 terms, the sum is -180, for the third A.P., 0.6, 1.7, 2.8,...... , to 100 terms, the sum is 5505, for the fourth A.P., 1/15 , 1/12 , 1/10 ,...., to 11 terms the sum is 33/20.

**☛ Related Questions:**

- 17th term of an AP exceeds its 10th term by 7. Find the common difference.
- Which term of the AP 3,15,27,39... will be 132 more than its 54th term?
- Two APs have the same common difference. The difference between their 100th term is 100, what is the difference between their 1000th terms?
- How many three-digit numbers are divisible by 7?

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