# 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 Sₙ = n/2 [2a + (n - 1) d] or Sₙ = n/2 [a + l], and the nth term of an AP is aₙ = 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,

- 2
^{nd}term, a₂ = 14 - 3
^{rd}term, a₃= 18 - Common difference, d = a₃ - a₂ = 18 - 14 = 4

We know that n^{th} term of an AP is, aₙ = a + (n - 1)d

a₂ = a + d

14 = a + 4

a = 10

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

S₅₁_{ }= 51/2 [2 × 10 + (51 - 1) 4]

= 51/2 [20 + 50 × 4]

= 51/2 × 220

= 51 × 110

= 5610

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

**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

**Summary:**

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

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