[tex]\bf \begin{array}{|ll|ll} \cline{1-2} term&value\\ \cline{1-2} a_3&19\\ a_4&19+d\\ a_5&19+d+d\\ a_6&19+d+d+d\\ &19+3d\\ &\\ \cline{1-2} \end{array}~\hspace{5em} \begin{array}{llll} \stackrel{a_6}{37}=19+3d\implies 18=3d \\\\\\ \cfrac{18}{3}=d\implies 6=d \end{array}[/tex]
since we know the common difference is d is 6, then to get from the 6th term to the 10th term, we use d 4 more times.
6th term..................37
7th term...................37 + 6
8th term...................37 + 6 + 6
9th term...................37 + 6 + 6 + 6
10th term...................37 + 6 + 6 + 6 + 6
.....................................61.
Answer: The tenth term is 61
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Work Shown:
third term = 19
fourth term = (third term)+d = 19+d for some number d
fifth term = (fourth term)+d = (19+d)+d = 19+2d
sixth term = (fifth term)+d = (19+2d)+d = 19+3d
sixth term = 37
Equate 19+3d and 37, then isolate d
19+3d = 37
3d = 37-19
3d = 18
d = 18/3
d = 6
Common difference is 6. Each time we want a new term, we add on 6
So,
seventh term = (sixth term) + d = 37 + d = 37+6 = 43
eighth term = (seventh term) + d = 43 + d = 43 + 6 = 49
ninth term = (eighth term) + d = 49+d = 49+6 = 55
tenth term = (ninth term) + d = 55+d = 55+6 = 61
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A shortcut is to use the formula
a_n = a_1 + d(n-1)
We found d = 6 earlier. We just need the first term a_1. Plug in d = 6 and n = 3
a_n = a_1 + d(n-1)
a_3 = a_1 + 6(3-1)
a_3 = a_1 + 12
Now replace a_3 with 19 which is the third term
19 = a_1 + 12
Isolate a_1 by subtracting 12 from both sides
19-12 = a_1 + 12-12
7 = a_1
a_1 = 7
Therefore the nth term formula is
a_n = 7 + 6(n-1)
We can plug in n = 10 to compute the following
a_10 = 7 + 6(10-1)
a_10 = 7 + 6(9)
a_10 = 7 + 54
a_10 = 61
we get the same answer as before.
