Question

In an A.P. 1st term is 1 and the last term is 20. The sum of all terms is = 399 then n = . . .
(A) 42
(B) 38
(C) 21
(D) 19

Answer 1

Answer:

Option B   n = 38 is correct

Step-by-step explanation:

Sum of an AP series is given by formula

S = (n/2)(a + L)

a = 1st term

L = Last Term

n = Number of Terms

S = Sum

399 = (n/2)(1 + 20)

=> n = 399 * 2 / 21

=> n = 19 * 2

=> n = 38

Option B n = 38 is correct

Answer 2

Answer:

38

Step-by-step explanation:

Given In an A.P. 1st term is 1 and the last term is 20. The sum of all terms is = 399 then n = . . .

We are given Sn = 399, aₙ = 20, a = 1

So we know the formula of sum to n terms

Sₙ = n/2(2 a + (n - 1)d)

 2n + n(n - 1)d = 798--------1

 we know aₙ = 20

a + (n - 1)d = 20

1 + (n - 1)d = 20

(n - 1)d = 19----------2

So substituting in 1 we get

2n + 19n = 798

21 n = 798

 n = 798 / 21

n = 38