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

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, $n=38$.

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Updated on: 10-Oct-2022

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