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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Updated on: 10-Oct-2022

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