# What is the 20th term of the arithmetic sequence 2, –4, –10, … ?

Mar 5, 2018

$- 112$

#### Explanation:

the nth term of an AP

$a , a + d , a + 2 d , . .$

is ${n}^{t h} = a + \left(n - 1\right) d$

here we have

$2 , - 4 , - 10 , \ldots$

$a = 2$

and it is seen that $d = - 6$

$\therefore {20}^{t h} = 2 + \left(20 - 1\right) \left(- 6\right)$

$= 2 + 19 \times - 6 = 2 - 114$

$= - 112$

Mar 5, 2018

${20}^{t h}$ term $= - 112$

#### Explanation:

If the sequence is denoted by the series ${a}_{i}$
then ${a}_{i} = {a}_{i - 1} - 6$

Setting ${a}_{0} = 8$
so that the first term is ${a}_{1} = 2$ (as given)

we have ${a}_{n} = {a}_{0} - \left(n \cdot 6\right)$

For $n = 20$
$\textcolor{w h i t e}{\text{XXX}} {a}_{20} = 8 - 20 \cdot 6 = 8 - 120 = - 112$

Mar 5, 2018

${T}_{20} = - 112$

#### Explanation:

The terms in the sequence $2 , - 4 , - 10 \ldots$ all differ by $- 6$

You can keep writing the terms until you get to the 20th, but that is not a very mathematical way of doing it and would be VERY time-consuming if you were asked for the 200th term for example.

Find the rule for the sequence... called ${T}_{n}$

If the difference is $- 6$, then the rule starts with ${T}_{n} = - 6 n$

You need to adjust the rule to start at the correct number,
$n$ starts from $1 , 2 , 3 , 4 \ldots$ for each successive term.

When $n = 1 , \text{ } - 6 \left(1\right) = - 6$

But the first term must be $2$ therefore add $8$

${T}_{n} = - 6 n + 8$

Check if $n = 2$
${T}_{2} = - 6 \left(2\right) + 8 = - 12 + 8 = - 4$

Check if $n = 3$
${T}_{3} = - 6 \left(3\right) + 8 = - 18 + 8 = - 10$

Now find the 20th term:

${T}_{20} = - 6 \left(20\right) + 8$

$= - 120 + 8 = - 112$