The correct answer is A. NGNGA
To find a sequence is arithmetic, we are going to find its common difference: [tex]d=a_{n}-a_{n-1}[/tex]
where
[tex]d[/tex] is the common difference
[tex]a_{n}[/tex] is the current term in the sequence
[tex]a_{n-1}[/tex] is the previous term in the sequence
To find if a sequence is geometric, we are going to find its common ratio:
[tex]r= \frac{a_{n}}{a_{n-1}} [/tex]
where
[tex]r[/tex] is the common ratio
[tex]a_{n}[/tex] is the current term in the sequence
[tex]a_{n-1}[/tex] is the previous term in the sequence
1) 3,4,6,10,18
For [tex]a_{n}=4[/tex] and [tex]a_{n-1}=3[/tex]:
[tex]d=a_{n}-a_{n-1}[/tex]
[tex]d=4-3[/tex]
[tex]d=1[/tex]
For [tex]a_{n}=6[/tex] and [tex]a_{n-1}=4[/tex]:
[tex]d=6-4[/tex]
[tex]d=2[/tex]
For [tex]a_{n}=4[/tex] and [tex]a_{n-1}=3[/tex]:
[tex]r= \frac{a_{n}}{a_{n-1}} [/tex]
[tex]r= \frac{4}{3} [/tex]
For [tex]a_{n}=6[/tex] and [tex]a_{n-1}=4[/tex]:
[tex]r= \frac{6}{4} [/tex]
[tex]r= \frac{3}{2} [/tex]
We can conclude that the sequence is neither arithmetic nor geometric.
2) [tex] \frac{25}{4} , \frac{5}{2} ,1...[/tex]
For [tex]a_{n}= \frac{5}{2} [/tex] and [tex]a_{n-1}= \frac{25}{4} [/tex]:
[tex]d=a_{n}-a_{n-1}[/tex]
[tex]d=\frac{5}{2}-\frac{25}{4}[/tex]
[tex]d=- \frac{15}{4} [/tex]
For [tex]a_{n}=1[/tex] and [tex]a_{n-1}= \frac{5}{2} [/tex]:
[tex]d= 1-\frac{5}{2}[/tex]
[tex]d= -\frac{3}{2} [/tex]
For [tex]a_{n}= \frac{5}{2} [/tex] and [tex]a_{n-1}= \frac{25}{4} [/tex]:
[tex]r= \frac{a_{n}}{a_{n-1}} [/tex]
[tex]r= \frac{\frac{5}{2} }{\frac{25}{4} } [/tex]
[tex]r= \frac{2}{5} [/tex]
For [tex]a_{n}=1[/tex] and [tex]a_{n-1}= \frac{5}{2} [/tex]:
[tex]r= \frac{1}{\frac{5}{2}} [/tex]
[tex]r= \frac{2}{5} [/tex]
We have a common ratio, we can conclude that the sequence is geometric.
3) [tex] \frac{2}{3} , \frac{4}{6} , \frac{8}{9} ...[/tex]
For [tex]a_{n}=\frac{4}{6}[/tex] and [tex]a_{n-1}= \frac{2}{3}[/tex]:
[tex]d=\frac{4}{6}-\frac{2}{3}[/tex]
[tex]d=0[/tex]
For [tex]a_{n}=\frac{4}{6}[/tex] and [tex]a_{n-1}= \frac{2}{3}[/tex]:
[tex]r= \frac{\frac{4}{6}}{\frac{2}{3}} [/tex]
[tex]r=1[/tex]
For [tex]a_{n}=\frac{8}{9}[/tex] and [tex]a_{n-1}= \frac{4}{6}[/tex]:
[tex]r= \frac{\frac{8}{9}}{\frac{4}{6}} [/tex]
[tex]r= \frac{4}{3} [/tex]
We can conclude that the sequence is neither arithmetic nor geometric.
4) 3,15,75...
For [tex]a_{n}=15[/tex] and [tex]a_{n-1}=3[/tex]:
[tex]d=15-3[/tex]
[tex]d=12[/tex]
For [tex]a_{n}=75[/tex] and [tex]a_{n-1}=15[/tex]:
[tex]d=75-15[/tex]
[tex]d=60[/tex]
For [tex]a_{n}=15[/tex] and [tex]a_{n-1}=3[/tex]:
[tex]r= \frac{15}{3} [/tex]
[tex]r=5[/tex]
For [tex]a_{n}=75[/tex] and [tex]a_{n-1}=15[/tex]:
[tex]r= \frac{75}{15} [/tex]
[tex]r=5[/tex]
We have a common ratio, we can conclude that the sequence is geometric.
5) 5,-11,-27
For [tex]a_{n}=-11[/tex] and [tex]a_{n-1}=5[/tex]:
[tex]d=-11-5[/tex]
[tex]d=-16[/tex]
For [tex]a_{n}=-27[/tex] and [tex]a_{n-1}=-11[/tex]:
[tex]d=-27--11[/tex]
[tex]d=-27+11[/tex]
[tex]d=-16[/tex]
We have a common difference, we can conclude that the sequence is arithmetic.
