In an A.P. if $a = 1,\ a_n = 20$ and $S_n=399$, then find $n$.
Given: In an A.P. if $a = 1,\ a_n = 20$ and $S_n=399$.
To do: 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 known,
$a_n=a+( n-1)d$
$\Rightarrow 20=1+( n-1)d$
$\Rightarrow ( n-1)d=20-1=19\ .........\ ( i)$
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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