Question

If an AP has $a=1,a_n=20$ and $S_n=399$, then find the value of $n$.

Answer 1

Compute the value of $\mathit{n}$:

Formulae:

The $n^{\mathrm{th}}$ term of an AP is $a_n=a+\left(n-1\right)d$ and the sum of $n$ terms of a series in an AP is $S_n=\frac{n}{2}\left[2a+(n-1)d\right]$.

Substitute $a=1,a_n=20$ in the $n^{\mathrm{th}}$ term of an AP formula.

$\begin{array}{rcl}& \Rightarrow & a_n=a+\left(n-1\right)d\\ & \Rightarrow & 20=1+\left(n-1\right)d\\ & \Rightarrow & 19=\left(n-1\right)d\end{array}$

Substitute $a=1$, $\left(n-1\right)d=19$ and $S_n=399$ into the sum of $n$ terms of a series in an AP formula.

$$\begin{gathered} \Rightarrow S_n=\frac{n}{2}\left[2\left(1\right)+\left(n-1\right)d\right] \\ \Rightarrow 399=\frac{n}{2}\left(2+19\right) \\ \Rightarrow 798=21n \\ \Rightarrow n=38 \end{gathered}$$

Hence, the value of $n=38$.