Question

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

Answer 1

Sum of the first n terms of an AP is given by:
Sₙ = $\frac{n}{2}$ [2a + (n - 1) d] or Sₙ = $\frac{n}{2}$ [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,
2nd term, a₂ = 14
3rd term, a₃= 18
Common difference, d = a₃ - a₂ = 18 - 14 = 4
We know that nth term of an AP is, aₙ = a + (n - 1)d
a₂ = a + d
14 = a + 4
a = 10
Sₙ = $\frac{n}{2}$ [2a + (n - 1) d]

S₅₁ = $\frac{51}{2}$ [2 × 10 + (51 - 1) 4]

= $\frac{51}{2}$ [20 + 50 × 4]

= $\frac{51}{2}$ × 220

= 51 × 110
= 5610