**The**

*n*th Term Of An Arithmetic SequenceIf {

*a*} is the

_{n}*n*th term of an arithmetic sequence with common difference

*d*, then

*a*

_{2}=

*a*

_{1}+

*d*

*a*

_{3}=

*a*

_{2}+

*d*=

*a*

_{1}+2

*d*

*a*

_{4}=

*a*

_{3}+

*d*=

*a*

_{1}+3

*d*

*a*=

_{n}*a*

_{1}+(

*n*-1)

*d*for every

*n*>1.

If we know the first term in an arithmetic progression, and the difference between terms, then we can work out the *n*th term, i.e. we can work out what any term will be. The formula which tells us what the *n*th term in an arithmetic progression is

*a*=

_{n}*a*+(

*n*-1)

*d*

where

*a*is the first term.

If we use

afor the first term,dfor the common difference, then the general termafor an arithmetic sequence is:_{n}a=_{n}a+(n-1)d

Example 1. Check whether 301 is a term of the list of numbers 5, 11, 17, 23, …

Solution:

We have:

*a*

_{2}–

*a*

_{1}=11-5=6,

*a*

_{3}–

*a*

_{2}=17-11=6,

*a*

_{4}–

*a*

_{3}=23-17=6

As

*a*

_{(k+1)}–

*a*is the same for k=1, 2, 3, etc., the given list of numbers is an AP.

_{k}Now,

*a*=5 and

*d*=6.

Let 301 be a term, say, the

*n*th term of the this AP.

We know that

*a*=

_{n}*a*+(

*n*-1)

*d*

301=5+(

*n*-1)x6

301=6

*n*-1

*n*=302/6=151/3

But

*n*should be a positive integer (Why?). So, 301 is not a term of the given list of numbers.

Example 2. Check whether -150 is a term of the AP: 11, 8, 5, 2, …

Solution:

Here, *a*=11 and *d*=8-11=-3. Let, the *n*th term of the A.P. is -150. Therefore, *a _{n}*=-150

*a*+(

*n*-1)

*d*=-150

11+(

*n*-1)(-3)=-150

11-3

*n*+3 =-150

-3

*n*=-164

*n*=164/3=54⅔

Here,

*n*is not a natural number, therefore, -150 is not the term of A.P., 11, 8, 5, 2, …

Example 3. Janeth saved $5 in the first week of a year and then increased her weekly savings by $1.75. lf in the *n*th week, her weekly savings become $20.75, find *n*.

Solution:

Savings for the first week =*a*=$5 increment in savings =*d*=$1.75 Let, her saving become $20.75 after *n* weeks.

Therefore, *a _{n}*=20.75

*a*+(

*n*-1)

*d*=20.75

5+(

*n*-1)(1.75)=20.75

(

*n*-1)(1.75)=15.75

*n*-1=15.75/1.75=9

*n*=10

Hence, her saving become $20.75, after 10 weeks.

Example 4: Calculating Terms in a Given Arithmetic Sequence

For this arithmetic sequence: -3, 2, 7, 12, …

a) Determine *t*_{20}.

b) Which term in the sequence has the value 212?

Solution:

a) Calculate the common difference: 2-(-3)=5

Use:

*t*=

_{n}*t*

_{1}+

*d*(

*n*-1)

Substitute:

*n*=20,

*t*

_{1}=-3,

*d*=5

*t*

_{20}=-3+5(20-1)

Use the order of operations.

*t*

_{20}=-3+5(19)

*t*

_{20}=92

b) Use

*t*=

_{n}*t*

_{1}+

*d*(

*n*-1):

Substitute:

*t*=212,

_{n}*t*

_{1}=-3,

*d*=5

*n*-1)

Solve for

*n*.

212=-3+5

*n*-5

220=5

*n*

220/5=

*n*

*n*=44

The term with value 212 is

*t*

_{44}.

Ex5. Which term of the AP: 21, 18, 15, … is -81? Also, is any term 0? Give reason for your answer.

Solution:

Here, *a*=21, *d*=18-21=-3 and *a _{n}*=-81, and we have to find

*n*.

As

*a*=

_{n}*a*+(

*n*-1)

*d*,

*n*-1)(-3)

-81=24-3

*n*

-105=-3

*n*

So,

*n*=35

Therefore, the 35th term of the given AP is -81.

Next, we want to know if there is any

*n*for which

*a*=0. If such an

_{n}*n*is there, then

*n*-1)⋅(-3)=0,

3(

*n*-1)=21

*n*=8

So, the eighth term is 0.

Ex6. Consider this number pattern: 8, 5, 2, …

a) Calculate the 15th term.

b) Determine which term of this sequence is -289.

Solution:

a) It is an arithmetic sequence: *a*+8, *d*=5-8=2-5=-3

*a*=

_{n}*a*+(

*n*-1)

*d*

*a*

_{15}=8+(15-1)(-3)

*a*

_{15}=8+14(-3)

*a*

_{15}=8-42=-34

b)

*a*=

_{n}*a*+(

*n*-1)

*d*

*n*-1)(-3)

-289=8-3

*n*+3

-300=-3

*n*

100=

*n*

-289 will be the 100th term.

Ex7. Given the sequence 6, 13, 20, …

a) Determine a formula for the *n*th term of the sequence.

b) Calculate the 21st term of this sequence.

c) Determine which term of this sequence is 97.

Solution:

a) It is an arithmetic sequence because there is a common difference.

*a*=6,

*d*=7,

*a*=

_{n}*a*+(

*n*-1)

*d*

*a*=6+(

_{n}*n*-1)(7)

*a*=7

_{n}*n*-1

b)

*a*

_{21}=7(21)-1+147-1+146

c) 97+7

*n*-1

*n*

14+

*n*

∴ 97 is the 14th term of the sequence.

Ex8. Given the sequence 4, 10, 16, …

a) Determine a formula for the *n*th term of the sequence.

b) Calculate the 50th term.

c) Which term of the sequence is equal to 310

Solution:

a) *a*=4 and *d*=10-4=16-10=6

*a*=

_{n}*a*+(

*n*-1)

*d*

=4+(

*n*-1)6

=4+6

*n*-6

=6

*n*-2

or by looking at the structure, the numbers are 2 less than the multiples of 6 i.e

*a*=6

_{n}*n*-2.

b)

*a*

_{50}=6×50-2=300-2=298

c) 6*n*-2=310

*n*=312

*n*=52

Example 9: Using an Arithmetic Sequence to Model and Solve a Problem

Some comets are called periodic comets because they appear regularly in our solar system. The comet Kojima appears about every 7 years and was last seen in the year 2007. Halley’s comet appears about every 76 years and was last seen in 1986.

Determine whether both comets should appear in 3043.

solution:

The years in which each comet appears form an arithmetic sequence. The arithmetic sequence for Kojima has *t*_{1}=2007 and *d*=7. To determine whether Kojima should appear in 3043, determine whether 3043 is a term of its sequence.

*t*=

_{n}*t*

_{1}+

*d*(

*n*-1)

Substitute:

*t*=3043,

_{n}*t*

_{1}=2007,

*d*=7

*n*-1)

Solve for

*n*.

*n*

1043=7

*n*

149=

*n*

Since the year 3043 is the 149th term in the sequence, Kojima should appear in 3043.

The arithmetic sequence for Halley’s comet has

*t*

_{1}=1986 and

*d*=76. To determine whether Halley’s comet should appear in 3043, determine whether 3043 is a term of its sequence.

*t*=

_{n}*t*

_{1}+

*d*(

*n*-1)

Substitute:

*t*=3043,

_{n}*t*

_{1}=1986,

*d*=76

*n*-1)

Solve for

*n*.

3043=1910+76

*n*

1133=76

*n*

*n*=14.9078 …

Since

*n*is not a natural number, the year 3043 is not a term in the arithmetic sequence for Halley’s comet; so the comet will not appear in that year.

Ex10. Ryan is a cartoonist. His comic strip has just been bought by a newspaper, so he sends them the 28 comic strips he has drawn so far. Each week after the first he mails 3 more comic strips to the newspaper.

a) Find the total number of comic strips sent after 1, 2, 3, and 4 weeks.

b) Show that the total number of comic strips sent after *n* weeks forms an arithmetic sequence.

c) Find the number of comic strips sent after 15 weeks.

d) When does Ryan send his 120th comic strip?

solution:

a) Week 1: 28 comic strips

Week 2: 28+3=31 comic strips

Week 3: 31+3=34 comic strips

Week 4: 34+3=37 comic strips

b) Every week, Ryan sends 3 comic strips, so the difference between successive weeks is always 3. We have an arithmetic sequence with *u*_{1}=28 and common difference *d*=3.

c) *u _{n}*=

*u*

_{1}+(

*n*-1)

*d*

*n*-1)×3

=25+3

*n*

∴

*u*

_{15}=25+3×15=70

After 15 weeks Ryan has sent 70 comic strips.

d) We want to find

*n*such that

*u*=120

_{n}*n*=120

3

*n*=95

*n*=31⅔

Ryan sends the 120th comic strip in the 32nd week.

Question 1. Does the number 203 belong to the arithmetic sequence 3, 7, 11, . . .?

Solution:

Here *a*=3 and *d*=4, so *a _{n}*=3+(

*n*-1)×4=4

*n*-1. We set 4

*n*-1=203 and find that

*n*=51. Hence, 203 is the 51st term of the sequence.

Question 2. Which term of the A.P.: 3, 8, 13, 18, … is 78?

Solution:

Here, *a*=3 and *d*=8-3=5.

Let, *n*th term of the A.P. is 78.

Therefore, *a _{n}*=78

*a*+(

*n*-1)

*d*=78

3+(

*n*-1)(5)=78

(

*n*-1)(5)=75

*n*-1=15

*n*=16

Hence, 16th term of the A.P.: 3, 8, 13, 18, … is 78.

Question 3. Daniel started work in 1995 at an annual salary of $5000 and received an increment of $200 each year. In which year did his income reach $7000?

Solution:

Starting salary =*a*=$5000 annual increment (common difference) =*d*=$200.

Let, after *n* years, his salary become $7000.

Therefore, *a _{n}*=7000

*a*+(

*n*-1)

*d*=7000

5000+(

*n*-1)(200)=7000

(

*n*-1)(200)=2000

*n*-1=10

*n*=11

Hence, in 11th year his salary become $7000.

Question 4. If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?

Solution:

Here, *a*_{3}=4 and *a*_{9}=-8 to find *n*, where *a _{n}*=0.

Given that:

*a*

_{3}=

*a*+(3-1)

*d*=4

*a*+2

*d*=4

*a*=4-2

*d*…(1)

and

*a*

_{9}=-8

⇒

*a*+8

*d*=-8

Putting the value of a from equation (1), we get

*d*+8

*d*=-8

6

*d*=-12

*d*=-2

Putting the value of

*d*in equation (1), we get

*a*=4-2(-2)=8

Putting the values in

*a*=0, we get

_{n}*a*=

_{n}*a*+(

*n*-1)

*d*=0

⇒8+(

*n*-1)(-2)=0

*n*-1=4

*n*=5

Hence, the 5th term of this AP is zero.

Question 5. Which term of the AP: 3, 15, 27, 39, … will be 132 more than its 54th term?

Solution:

First term=3 and common difference =15-3=12

Let the *n*th term of AP: 3, 15, 27, 39, … will be 132 more than its 54th term.

Therefore, *a _{n}*=

*a*

_{54}+132

*a*+(

*n*-1)

*d*=

*a*+53

*d*+132

(

*n*-1)(12)=53×12+132

(

*n*-1)(12)=768

*n*-1=768/12=64

*n*=65

Hence, 65th term of the AP: 3, 15, 27, 39, … will be 132 more than its 54th term.

Question 6. For what value of *n*, are the *n*th terms of two APs: 63, 65, 67 … and 3, 10, 17 … equal?

Solution:

First term of first AP =*A*=63 and common difference =*D*=65-63=2.

Therefore, *A _{n}*=

*A*+(

*n*-1)

*D*=

*A*=63+(

_{n}*n*-1)2

First term of second AP =

*a*=3 and common difference =

*d*=10-3=7.

Therefore,

*a*=

_{n}*a*+(

*n*-1)

*d*⇒

*a*=3+(

_{n}*n*-1)7

According to question,

*A*=an

_{n}*n*-1)2=3+(

*n*-1)7

63+2

*n*-2=3+7

*n*-7

65=5

*n*

*n*=13

Hence, the 13th term of both the APs are equal.

Question 7. Which term of the AP: 121, 117, 113 … is its first negative term?

[Hint: Find *n* for *a _{n}*<0]
Solution:
Here,

*a*=121 and

*d*=117-121=-4.

Let,

*a*be the first negative term of this AP.

_{n}*a*<0

_{n}*a*+(

*n*-1)

*d*<0 121+(

*n*-1)(-4)<0 121-4

*n*+4<0 125<4

*n*

*n*>125/4→

*n*>31.25

*n*=32

Hence, 32nd term of the AP: 121, 117, 113 … is its first negative term.

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