Question

Question 17
In an AP, if a = 1,
$a_n=20$
$and{S}_n$ = 399,
then n is equal to :

A) 19
B) 21
C) 38
D) 42

Answer 1

The sum up to n terms of an A.P is given by,
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$\Rightarrow 399=\frac{n}{2}[2\times 1+(n-1\left)d\right]$
$\Rightarrow 798=2n+n(n-1)d...\left(i\right)$
$and{a}_n=20$
$\Rightarrow a+(n-1)d=20...\left(ii\right)[\because a_n=a+(n-1\left)d\right]$
Using eq.(ii) in eq.(i), we get,
798 = 2n + 19n
$\Rightarrow$ 798 = 21n
$\therefore n=\frac{798}{21}=38$