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

Question

Philoid · Accepted Answer

We have; a₂ = 14, a₃ = 18 and n = 51

We know,

a₃ – a₂ = 18 – 14 = 4

Therefore, d = 4

a₂ – a = 4

14 – a = 4

a = 10

Now, sum can be calculated as follows:

= 51*(10 + 100)

= 51*110

= 5610