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

Question

Philoid · Accepted Answer

Second term (a₂) = 14; Third term (a₃) = 18

a₂ = a+(2-1)d

14 = a+d ⇒ 1

a₃ = a+(3-1)d

18 = a+2d ⇒ 2

By subtracting both the equations

14 -18 = a+d –(a+2d)

-4 = -d

d = 4

By substituting d in the equation 1

a + d = 14

a + 4 = 14

a = 10

The sum of first 51 terms is given by (s_n) = [2a + (n-1) d]

S₅₁ = [2×10 + (51-1) 4]

= 5610