Question

Find the tenth term of an arithmetic sequence if the third term is 19 and the sixth term is 37.

Answer 1

$\bf \begin{array}{|ll|ll} \cline{1-2} term&value\\ \cline{1-2} a_3&19\\ a_4&19+d\\ a_5&19+d+d\\ a_6&19+d+d+d\\ &19+3d\\ &\\ \cline{1-2} \end{array}~\hspace{5em} \begin{array}{llll} \stackrel{a_6}{37}=19+3d\implies 18=3d \\\\\\ \cfrac{18}{3}=d\implies 6=d \end{array}$

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 2

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.