Question 550735
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The formula for all of them is a<sub>n</sub> = a<sub>1</sub> + (n-1)d

direction: write a rule for the nth term of the arithmetic sequence

30. d=4, a14=46

an = a1 + (n-1)d
a14 = a1 + (14-1)(4)
46 = a1 + 13(4)
46 = a1 + 52
-6 = a1

an = a1 + (n-1)d
an = -6 + (n-1)(4)
an = -6 + 4n - 4
an = 4n - 10

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31. d=-12, a1=80. 

an = a1 + (n-1)d
an = 80 + (n-1)(-12)
an = 80 - 12n + 12
an = 92 - 12n

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32. d=5/3, a8=24

an = a1 + (n-1)d
a8 = a1 + (8-1)({{{5/3}}})
24 = a1 + (7)({{{5/3}}})
24 = a1 + {{{35/3}}}
3·24 = 3a1 + 35
72 = 3a1 + 35
37 = 3a1
{{{37/3}}} = a1

an = a1 + (n-1)d
an = {{{37/3}}} + (n-1)({{{5/3}}})

3an = 37 + (n-1)5
3an = 37 + 5n - 5
3an = 32 + 5n
 an = {{{(32+5n)/3}}}

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33. a5=17, a15=77

an = a1 + (n-1)d
a5 = a1 + (5-1)d
17 = a1 + 4d

an = a1 + (n-1)d
a15 = a1 + (15-1)d
77 = a1 + 14d

Solve system of equations:

{{{system(a1+4d=17,a1+14d=77)}}}

get a1 = -7, d = 6

an = a1 + (n-1)d
an = -7 + (n-1)(6)
an = -7 + 6n - 6
an = 6n - 13

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34. d=-6, a12=-4

an = a1 + (n-1)d
a12 = a1 + (12-1)(-6)
-4 = a1 + (11)(-6)
-4 = a1 - 66
62 = a1

an = a1 + (n-1)d
an = 62 + (n-1)(-6)
an = 62 - 6(n-1)
an = 62 - 6n + 6
an = 68 - 6n

You do the last two.  They're like 33.

35. a2=-28, a20=52
37. a7=34, a18=122

Edwin</pre>