Question

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

Answer 1

If a₂ = 14, a₃ = 18, then S₁₁ =? 

a₂ = 14a + d = 14 …………. (i) 

a₃ = 18a + 2d = 18 …………… (ii) 

from equation (1) – equation (11),

d =4. 

Substituting the value of ‘d’ In eqn. (i) 

a + d = 14 

a + 4 = 14 

a = 14 – 4 

∴ a = 10 

a = 10, d = 4

= 51 × 110 

∴ S₅₁ = 5610