Question

How do you find the ${12}^{\mathrm{th}}$ term of the arithmetic sequence$20,14,8,2,-4,...$?

Answer 1

Step-1: Find the common difference $d$:

The given series is $20,14,8,2,-4,...$

From the given series it can be observed that the first term $\left(a\right)$ and the second term $\left(a_2\right)$ are $20$ and $14$ respectively.

Subtract the first term of the series from the second term to obtain the common difference.

$\begin{array}{rcl}d& =& a_2-a\\ & \Rightarrow & d=14-20\\ \therefore d& =& -6\end{array}$

Thus, the common difference of the given arithmetic sequence is $d=-6$.

Step-2: Find the ${12}^{\mathrm{th}}$ term of the given series.

The $n^{\mathrm{th}}$ term of an arithmetic series is obtained by the formula $a_n=a+\left(n-1\right)d$,

where $a$ is the first term and $d$ is the common difference.

Put $n=12$,$d=-6$,$a=20$ in $a_n=a+\left(n-1\right)d$ .

$$\begin{gathered} a_{12}=20+\left(12-1\right)\left(-6\right) \\ \Rightarrow a_{12}=20+\left(11\right)\left(-6\right) \\ \Rightarrow a_{12}=20-66 \\ \therefore a_{12}=-46 \end{gathered}$$

Hence, the ${12}^{\mathrm{th}}$ term of the arithmetic series is $-46$.