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Question:

If $1+4+7+10+\ldots+x=287$, find the value of $x$.

 

Solution:

$s_{n}=\frac{n}{2}[2 a+(n-1) d]$

$a=1, d=4-1=3$ and $s_{n}=287$

$287=\frac{n}{2}[2+(n-1) 3]$

$\Rightarrow 574=n(2+3 n-3)$

$\Rightarrow 3 n^{2}-n-574=0$

Using $n=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$n=\frac{-(-1) \pm \sqrt{(-1)^{2}-4(3)(-574)}}{2(3)}$

$\Rightarrow n=\frac{1 \pm \sqrt{1+6888}}{6}$

$\Rightarrow n=14, \frac{-41}{3}$

Number of terms can not be negative

So, $n=14$

$n=14$

$s_{n}=\frac{n}{2}[a+l]$

$287=\frac{14}{2}[1+x]$

$\Rightarrow x=40$

 

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