Question

how many terms of A.P 30,48,66 must be taken to get the sum 4020​

Answer 1

Step-by-step explanation:

a = 30

d = 48 - 30

d =18

Sn = 4020

Sn = n/2(2a + (n-1)d)

4020 = n/2 (60 + (n-1)18)

8040 = 60n +18n^2 - 18n

42n + 18n^2 - 8040 = 0

9n^2 + 21n - 4020 = 0

3(3n^2 +7n-1340) = 0

3n^2 +67n -60n -1340 = 0

(3n - 60) (3n + 67)

n = 20 / - 67/3

since the terms can't be negative so

n = 20