Question

find the sum of first 51 terms of an ap whose second and third terms are 14 and 18 respectively.​

Answer 1

Answer:

Sum of first 51 terms = 5610

Step-by-step explanation:

Given:

2nd term, a₂ = 14

3rd term, a₃= 18

Common difference, d = a₃ - a₂ = 18 - 14 = 4

a₂ = a + d

14 = a + 4

a = 10

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

Answer 2

Answer:

5500

Step-by-step explanation:

a2 = 14

a3 = 18

a3 = a+2d

18=a+2d. i

a2=a+d

14=a+d. ii

sub I and ii

18= a+2d

-14=-a-d

4 = d

18 = a+8

a=10

s51 = 51/2 (20 + 200)

s51 = 51/2 x 220

= 51 x 110

= 5500