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B. Finding the n th Term of an Arithmetic Sequence

B. Finding the n th Term of an Arithmetic Sequence

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Section 9.2 Arithmetic Sequences



EXAMPLE 4







Determining the Number of Terms in an Arithmetic Sequence

Find the number of terms in the arithmetic sequence 2, Ϫ5, Ϫ12, Ϫ19, p , Ϫ411.



Solution







By inspection we see that a1 ϭ 2 and d ϭ Ϫ7. As before,

an ϭ a1 ϩ 1n Ϫ 12d

ϭ 2 ϩ 1n Ϫ 12 1Ϫ72

ϭ 2 Ϫ 7n ϩ 7

ϭ Ϫ7n ϩ 9



n th term formula

substitute 2 for a1 and Ϫ7 for d

distribute Ϫ7

simplify



Although we don’t know the number of terms in the sequence, we do know the last

or nth term is Ϫ411. Substituting Ϫ411 for an gives

Ϫ411 ϭ Ϫ7n ϩ 9

60 ϭ n



substitute Ϫ411 for an

solve for n



There are 60 terms in this sequence.

Now try Exercises 43 through 50







Note that in both Examples 3 and 4, the nth term had the form of a linear equation

1y ϭ mx ϩ b2 after simplifying: an ϭ 0.3n Ϫ 0.2 and an ϭ Ϫ7n ϩ 9. This is a characteristic of arithmetic sequences, with the common difference d corresponding to

the slope m. This means the graph of an arithmetic sequence will always be a set of

discrete points that lie on a straight line. After entering u1n2 ϭ 0.3n Ϫ 0.2 (from

Example 3), Figure 9.23 shows the table of values for an ϭ 0.3n Ϫ 0.2, with the graph

in Figure 9.24A.

Figure 9.24A



Figure 9.23



5



10



0



Ϫ1.5



Figure 9.24B

In sequence MODE , we still set the size of the viewing

window as before, but we can also stipulate the range

of values of n to be used (nMin and nMax), which

term we want to plot as a beginning (PlotStart), and

whether we want all following terms graphed (PlotStep ϭ 1), every second term graphed (PlotStep ϭ 2),

and so on. The graph in Figure 9.24A was generated using the values shown in Figure 9.24B

(Ymin ϭ Ϫ1.5, Ymax ϭ 5, and Yscl ϭ 1 cannot be seen).

To see the graph of a sequence more distinctly, we can enter the natural numbers 1

through 10 in L1, then define L2 as u(L1) (Figures 9.25 and 9.26) and plot these points

(Figure 9.27).



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Figure 9.27

Figure 9.25



Figure 9.26



5



10



0



Ϫ1.5



One additional advantage of this approach is that we can easily verify the common difference is d using the “ ¢ List(” feature, which automatically computes the difference

between each successive term in a specified list. This option is located in the same submenu as the “seq(” option, accessed using 2nd STAT (LIST)

(OPS) 7: ¢ List(L2)

. See Figures 9.28 and 9.29.

ENTER



Figure 9.28



EXAMPLE 5







Figure 9.29



Graphing Arithmetic Sequences

Enter the natural numbers 1 through 6 in L1, and the terms of the sequence Ϫ0.45,

Ϫ0.1, 0.25, 0.6, 0.95, 1.3 in L2. Then verify the sequence in L2 is arithmetic by

a. Graphing the related points to see if they appear linear.

b. Using the ¢ List( feature.

c. Finding the nth term that defines the sequence and graph the sequence.



Solution







a. The plotted points are shown in Figure 9.30 and appear to be linear.

b. The ¢ List( feature shows there is a common difference of 0.35 (Figure 9.31).

c. With a1 ϭ Ϫ0.45 and d ϭ 0.35, the nth term must be

an ϭ a1 ϩ 1n Ϫ 12d

ϭ Ϫ0.45 ϩ 1n Ϫ 12 10.352

ϭ 0.35n Ϫ 0.8



n th term formula

substitute for a1 and d

simplify



The nth term for this sequence is an ϭ 0.35n Ϫ 0.8. The graph is shown in

Figure 9.32.

Figure 9.32

Figure 9.30

Figure 9.31

2

2



7



0



Ϫ1



7



0



Ϫ1



Now try Exercises 51 through 54







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If the term a1 is unknown but a term ak is given, the nth term can be written

an ϭ ak ϩ 1n Ϫ k2d



(the subscript of the term ak and coefficient of d sum to n).

EXAMPLE 6







Finding the First Term of an Arithmetic Sequence

Given an arithmetic sequence where a6 ϭ 0.55 and a13 ϭ 0.9, find the common

difference d and the value of a1.



Solution







At first it seems that not enough information is given, but recall we can express a13

as the sum of any earlier term and the appropriate multiple of d. Since a6 is known,

we write a13 ϭ a6 ϩ 7d (note 13 ϭ 6 ϩ 7 as required).

a1 is unknown

a13 ϭ a6 ϩ 7d

0.9 ϭ 0.55 ϩ 7d substitute 0.9 for a13 and 0.55 for a6

0.35 ϭ 7d

subtract 0.55

d ϭ 0.05

solve for d

Having found d, we can now solve for a1.

a13 ϭ a1 ϩ 12d

0.9 ϭ a1 ϩ 1210.052

0.9 ϭ a1 ϩ 0.6

a1 ϭ 0.3



B. You’ve just seen how

we can find the nth term of an

arithmetic sequence



n th term formula for n ϭ 13

substitute 0.9 for a13 and 0.05 for d

simplify

solve for a1



The first term is a1 ϭ 0.3 and the common difference is d ϭ 0.05.

Now try Exercises 55 through 60







C. Finding the n th Partial Sum of an Arithmetic Sequence

Using sequences and series to solve applications often requires computing the sum of

a given number of terms. To develop and understand the approach used, consider the

sum of the first 10 natural numbers. Using S10 to represent this sum, we have

S10 ϭ 1 ϩ 2 ϩ 3 ϩ 4 ϩ 5 ϩ 6 ϩ 7 ϩ 8 ϩ 9 ϩ 10. We could just use brute force,

but if we rewrite the sum a second time but in reverse order, then add it to the first, we

find that each column adds to 11.

S10 ϭ 1 ϩ 2 ϩ 3 ϩ 4 ϩ 5 ϩ 6 ϩ 7 ϩ 8 ϩ 9 ϩ 10

S10 ϭ 10 ϩ 9 ϩ 8 ϩ 7 ϩ 6 ϩ 5 ϩ 4 ϩ 3 ϩ 2 ϩ 1

2S10 ϭ 11 ϩ 11 ϩ 11 ϩ 11 ϩ 11 ϩ 11 ϩ 11 ϩ 11 ϩ 11 ϩ 11

Since there are 10 columns, the total is 11 ϫ 10 ϭ 110 but this is twice the actual sum

and we find that 2S10 ϭ 110, so S10 ϭ 55.

Now consider the sequence a1, a2, a3, a4, p , an with common difference d. Use Sn

to represent the sum of the first n terms and write the original series, then the series in

reverse order underneath. Since one row increases at the same rate the other decreases,

the sum of each column remains constant, and for simplicity’s sake we choose a1 ϩ an

to represent this sum.

Sn ϭ

a1

ϩ

a2

ϩ

a3

ϩ p ϩ anϪ2 ϩ anϪ1 ϩ

an

add

p

columns

Sn ϭ

an

ϩ anϪ1 ϩ anϪ2 ϩ

ϩ

a3

ϩ

a2

ϩ

a1

p

2Sn ϭ 1a1 ϩ an 2 ϩ 1a1 ϩ an 2 ϩ 1a1 ϩ an 2 ϩ

ϩ 1a1 ϩ an 2 ϩ 1a1 ϩ an 2 ϩ 1a1 ϩ an 2 ↓ vertically

To understand why each column adds to a1 ϩ an, consider the sum in the second

column: a2 ϩ anϪ1. From a2 ϭ a1 ϩ d and anϪ1 ϭ an Ϫ d, we obtain a2 ϩ anϪ1 ϭ

1a1 ϩ d2 ϩ 1an Ϫ d2 by adding the equations, which gives a result of a1 ϩ an. Since

there are n columns, we end up with 2Sn ϭ n1a1 ϩ an 2, and solving for Sn gives the

formula for the first n terms of an arithmetic sequence.



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The nth Partial Sum of an Arithmetic Sequence

Given an arithmetic sequence with first term a1, the nth partial sum is given by

Sn ϭ



n

1a1 ϩ an 2.

2



In words: The sum of an arithmetic sequence is one-half the number of terms times the

sum of the first and last term.



EXAMPLE 7







Computing the Sum of an Arithmetic Sequence

Use the summation formula to find the sum of the first 75 positive odd integers:

75



͚ 12n Ϫ 12 . Verify the result using a graphing calculator.



nϭ1



Solution







The initial terms of the sequence are 1, 3, 5, p and we note a1 ϭ 1, d ϭ 2, and

n ϭ 75. To use the sum formula, we need the value of a75: 21752 Ϫ 1 ϭ 149.

formula shows a75 ϭ a1 ϩ 74d ϭ 1 ϩ 74122, so a75 ϭ 149.

n

Sn ϭ 1a1 ϩ an 2

2

75

S75 ϭ 1a1 ϩ a75 2

2

75

ϭ 11 ϩ 1492

2

ϭ 5625



sum formula



substitute 75 for n



substitute 1 for a1, 149 for a75

result



The sum of the first 75 positive odd integers is 5625.

To verify, we enter u1n2 ϭ 2n Ϫ 1 on the Y=

screen, and find the sum of the first 75 terms of the

sequence on the home screen as before. See figure.



Now try Exercises 61 through 66







By substituting the nth term formula directly into the formula for partial sums,

we’re able to find a partial sum without actually having to find the nth term:



C. You’ve just seen how

we can find the nth partial sum

of an arithmetic sequence



n

Sn ϭ 1a1 ϩ an 2

2

n

ϭ 1a1 ϩ 3a1 ϩ 1n Ϫ 12d 4 2

2

n

ϭ 3 2a1 ϩ 1n Ϫ 12d 4

2



sum formula

substitute a1 ϩ 1n Ϫ 12d for an



alternative formula for the nth partial sum



See Exercises 67 through 72 for more on this alternative formula.



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