Question

Please answer this for me

Answer 1

In 1-4, to determine whether a sequence is either arithmetic or geometric, you need to look at differences of consecutive terms (arithmetic) and ratios of consecutive terms (geometric). If you can't find it, the sequence will fall under the "neither" category.

For example, the differences between consecutive terms in the first sequence are

$\left\{2-4,\dfrac12-2,\dfrac14-\dfrac12,\ldots\right\}=\left\{-2,-\dfrac32,-\dfrac14,\ldots\right\}$

If the sequence was arithmetic, the difference between consecutive terms would have been the same constant throughout this list. But that's not the case, so this sequence is not arithmetic.

The ratios between consecutive terms are

$\left\{\dfrac24,\dfrac{\frac12}2,\dfrac{\frac14}{\frac12},\ldots\right\}=\left\{\dfrac12,\dfrac14,\dfrac12,\ldots\right\}$

The sequence would have been geometric if the list contained the same value throughout, but it doesn't. So this sequence is neither arithmetic nor geometric.

Meanwhile, in the second sequence, the differences are

$\{-1-(-6),4-(-1),9-4,\ldots\}=\{5,5,5,\ldots\}$

so this sequence is arithmetic.

In 5-6, you know the sequences are arithmetic, so you know that they follow the recursive rule

$a_n=a_{n-1}+d$

For example, in the fifth sequence we know the first term is $a_1=4$. The common difference between terms is $d=9-4=5$. So using the rule above, we have the pattern

$a_2=a_1+d$

$a_3=a_2+d=a_1+d(2)$

$a_4=a_3+d=a_1+d(3)$

and so on, so that the $n$-th term is determined entirely by $a_1$ with the formula

$a_n=a_1+d(n-1)$

This means the 21st term in the fifth sequence is

$a_{21}=a_1+5(21-1)=4+5(20)=104$

The process is simple: identify $a_1$ and $d$, plug them into the formula above, then evaluate it at whatever $n$ you need to use.