# Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively

**Solution:**

Sum of the first n terms of an AP is given by S_{n} = n/2 [2a + (n - 1) d] or S_{n} = n/2 [a + l], and the nth term of an AP is a_{n} = a + (n - 1)d

Here, a is the first term, d is the common difference and n is the number of terms and l is the last term.

Given,

- 2nd term, a
_{2}= 14 - 3rd term, a
_{3}= 18 - Common difference, d = a
_{3}- a_{2}=18 - 14 = 4

We know that nth term of AP, a_{n} = a + (n - 1)d

a_{2} = a + d

14 = a + 4

a = 10

Sum of n terms of AP is given by S_{n} = n / 2 [2a + (n - 1) d]

S_{51 }= 51 / 2 [2 × 10 + (51 - 1) 4]

= 51 / 2 [20 + 50 × 4]

= 51 / 2 × 220

= 51 × 110

= 5610

**Video Solution:**

## Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively

### Class 10 Maths NCERT Solutions - Chapter 5 Exercise 5.3 Question 8:

Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively

The sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively is 5610