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If $1+2+3+........+n=78$, then find the value of $n$.
Given: An A.P. $1+2+3+........+n=78$
To do: To find the value of $n$.
Solution:
As given $1+2+3+........+n=78$
Here $a=1$, $d=1$, sum of $n$ terms of the A.P., $S_n=78$ and number of terms$=n$.
As known $S_n=\frac{n}{2}[2a+( n-1)d]$
On substituting the values,
$78=\frac{n}{2}[2\times1+( n-1)1]$
$\Rightarrow \frac{n}{2}[n+1]=78$
$\Rightarrow \frac{n^2+n}{2}=78$
$\Rightarrow n^2+n=156$
$\Rightarrow n^2+n-156=0$
$\Rightarrow n^2+13n-12n-156=0$
$\Rightarrow n( n+13)-12( n+13)=0$
$\Rightarrow ( n-12)( n+13)=0$
$\Rightarrow n=12,\ n=-13$
$\because\ n$ is a natural number, it can't be negative.
$\therefore\ n=12$
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