Find an example of an arithmetic sequence that has an explicit pattern. Then determine the formula for that sequence and use it to define the given term.

Find an example of an arithmetic sequence that has an explicit pattern Then determine the formula for that sequence and use it to define the given term class=

Respuesta :

11)

5, -5, -15, -25,  ...

-5-5 = -10;   -15-(-5) = -10;   -25-(-15) = -10   ⇒  this is an example of arithmetic sequence that has explicit patern (evey next term is 10 less than previous one)

difference:  d = -10

first term:  a = 5

so the formula:

                         [tex]a_n=a+d(n-1)\\\\a_n=5+(-10)(n-1)\\\\a_n =5-10n+10\\\\\underline{a_n=-10n+15}[/tex]                        

and:

       [tex]a_{20}=-10\cdot20+15=-200+15=-185[/tex]

12)

19, 26, 33, 40,  ...

26-19 = 7;   33-26 = 7;   40-33 = 7   ⇒  this is an example of arithmetic sequence that has explicit patern (evey next term is 7 more than previous one)

difference:  d = 7

first term:  a = 19

so the formula:

                         [tex]a_n=a+d(n-1)\\\\a_n=19+7(n-1)\\\\a_n =19+7n-7\\\\\underline{a_n=7n+12}[/tex]                        

and:

      [tex]a_{39}=7\cdot39+12=273+12=285[/tex]

13)

-20, -29, -38, -47,  ...

-29-(-20) = -9;   -38-(-28) = -9;   -47-(-38) = -9   ⇒  this is an example of arithmetic sequence that has explicit patern (evey next term is 9 less than previous one)

difference:  d = -9

first term:  a = -20

so the formula:

                         [tex]a_n=a+d(n-1)\\\\a_n=-20+(-9)(n-1)\\\\a_n =-20-9n+9\\\\\underline{a_n=-9n-11}[/tex]                        

and:

       [tex]a_{12}=-9\cdot12-11=-108-11=-119[/tex]

11)

1, -2, 3, -4,  ...

-5-1 = -3;   3-(-2) = 5 ≠ -3   ⇒  this is NOT an example of arithmetic sequence

however it has explicit patern:

                         [tex]a_1=1 =1\cdot\left(-1\right)^0= 1\cdot\left(-1\right)^{1-1}\\\\a_2=-2=2\cdot\left(-1\right)^1=2\cdot\left(-1\right)^{2-1}\\\\a_3=3=3\cdot\left(-1\right)^2=3\cdot\left(-1\right)^{3-1}\\\\a_4=-4=4\cdot\left(-1\right)^3=4\cdot\left(-1\right)^{4-1}\\\\\underline{a_n=n\cdot\left(-1\right)^{n-1}}[/tex]                        

so:

     [tex]a_{33}=33\cdot\left(-1\right)^{33-1}=33\cdot\left(-1\right)^{32}=33\cdot1=33[/tex]