Question

In an AP $a=5,d=3$, $a_n=50$ find $n$ and $S_n$

Answer 1

We have,

$a=5,d=3,a_n=50$

We know that

$a_n=a+(n-1)d$

So,

$50=5+(n-1)3$

$(n-1)3=45$

$n-1=15$

$n=16$

We know that

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

Therefore,

$S_{16}=\frac{16}{2}[2\times 5+(16-1\left)3\right]$

$S_{16}=8[10+15\times 3]$

$S_{16}=8[10+45]$

$S_{16}=8\left[55\right]$

$S_{16}=440$

Hence, this is the answer.