Question

Determine if the sequence is arithmetic. If it is, find the common difference. Is the sequence a function ?

Answer 1

$6)\\r=a_{n+1}-a_n\to r=a_2-a_1=a_3-a_2=a_4-a_3=...\\\\a_1=-7;\ a_2=-10;\ a_3=-11;\ a_4=-13\\\\a_2-a_1=-10-(-7)=-10+7=-3\\a_3-a_2=-11-(-10)=-11+10=-1\\\\a_3-a_2\neq a_2-a_1\\\\\text{It's not an arithmetic sequence}$


$\r=\dfrac{a_{n+1}}{a_n}\\\\r=\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=\dfrac{a_4}{a_3}=...\\\\8.)\\a_1=4;\ a_2=20;\ a_3=100;\ a_4=500\\\\r=\dfrac{20}{4}=\dfrac{100}{20}=\dfrac{500}{100}=5\\\\10.)\\a_1-30;\ a_2=-45;\ a_3=-50;\ a_4=-65\\\\\dfrac{a_2}{a_1}=\dfrac{-45}{-30}=\dfrac{3}{2}\\\\\dfrac{a_3}{a_2}=\dfrac{-50}{-45}=\dfrac{10}{9}\\\\\dfrac{a_2}{a_1}\neq\dfrac{a_3}{a_2}\\\\\text{It's not a geometric sequence}$
$12.)\\a_1=-7;\ a_2=-14;\ a_3=-21;\ a_4=-28\\\\\dfrac{a_2}{a_1}=\dfrac{-14}{-7}=2\\\\\dfrac{a_3}{a_2}=\dfrac{-21}{-14}=\dfrac{3}{2}\\\\\dfrac{a_2}{a_1}\neq\dfrac{a_3}{a_2}\\\\\text{It's not a geometric sequence}\\\\\text{It's a function}$