Question

find the 20 terms of ap whose second term is 14 and third ​

Answer 1

Answer:

common difference (d) = 18-14 = 4

the second term = a+d = 14

so, a = 14-4 = 10

SUM

number of terms (n) = 51

a = 10

d = 4

Sum=n/2 [2a + (n-1)d]

       = 51/2 [20 + (50 x 4)]

       = 51/2 [20+200]

       = 51 x 110

       = 5610

therefore, the sum of 51 terms of AP is 5610