# Choose the correct answer from the given four options:In an AP if $a=1, a_{n}=20$ and $S_{n}=399$, then $n$ is(A) 19(B) 21(C) 38(D) 42

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Given:

In an A.P., $a = 1,\ a_n = 20$ and $S_n=399$.

To do:

We have to find the value of $n$.

Solution:

Here, $a = 1,\ a_n = 20$ and $S_n=399$

Let $d$ be the common difference of the A.P.

As we know,

$a_n=a+( n-1)d$

$\Rightarrow 20=1+( n-1)d$

$\Rightarrow ( n-1)d=20-1=19\ .........\ ( i)$

Sum of the $n$ terms of the A.P., $S_n=\frac{n}{2}[2a+( n-1)d]$

$\Rightarrow 399=\frac{n}{2}[2\times1+19]$

$\Rightarrow 399=\frac{n}{2}[21]$

$\Rightarrow n=\frac{399\times2}{21}$

$\Rightarrow n=38$

Thus, $n=38$.

Updated on 10-Oct-2022 13:27:27