**Arithmetic Sequences**

The simplest type of numerical sequence is an arithmetic sequence.

Definition: Arithmetic Sequence

Anarithmetic(orlinear) sequence is a sequence of numbers in which each new term is calculated byaddinga constant value to the previous term

For example, 1, 2, 3, 4, 5, 6, … is an arithmetic sequence because you add 1 to the current term to get the next term:

second term: 2=1+1

third term: 3=2+1

⋮

*n*th term:

*n*=(

*n*-1)+1

**General Equation for the nth-term of an Arithmetic Sequence**

More formally, the number we start out with is called

*a*

_{1}(the first term), and the difference between each successive term is denoted

*d*, called the

*common difference*.

The general arithmetic sequence looks like:

*a*

_{1}=

*a*

_{1}

*a*

_{2}=

*a*

_{1}+

*d*

*a*

_{3}=

*a*

_{2}+

*d*=(

*a*

_{1}+

*d*)+

*d*=

*a*

_{1}+2

*d*

*a*

_{4}=

*a*

_{3}+

*d*=(

*a*

_{1}+2

*d*)+

*d*=

*a*

_{1}+3d

⋮

*a*=

_{n}*a*

_{1}+

*d*⋅(

*n*-1)

Thus, the equation for the

*n*th-term will be:

*a*=

_{n}*a*

_{1}+

*d*⋅(

*n*-1)

Given

*a*

_{1}and the common difference,

*d*, the entire set of numbers belonging to an arithmetic sequence can be generated.

Definition: Arithmetic Sequence

Anarithmetic(orlinear) sequence is a sequence of numbers in which each new term is calculated by adding a constant value to the previous term:

a=_{n}a_{(n-1)}+d

where

●arepresents the new term, the_{n}nth-term, that is calculated;

●a_{(n-1)}represents the previous term, the (n-1)th-term;

●drepresents some constant.

**Arithmetic Progression (A.P.)**

An **arithmetic progression** is a sequence of terms such that the difference between any term and the one immediately preceding it is a constant. This difference is called the common difference.

For example, the sequences

(i) 3, 7, 11, 15, 19, …

(ii) 7, 5, 3, 1, -1, …

(iii) 1, ¾, ½, ¼, 0, -¼, …

(iv) 2, 2, 2, 2, …

are arithmetic progressions.

In (i) common difference is 4,

In (ii) common difference is -2,

In (iii) common difference is -¼, and

In (iv) common difference is 0,

In general, an arithmetic progression (A.P.) is given by

*a*,

*a*+

*d*,

*a*+2

*d*,

*a*+3

*d*, ….

We call

*a*as

**first term**and

*d*as the

**common difference**.

The

*n*th term of the above A.P. is denoted by an and is given by

*a*

_{n}*a*=

_{n}*a*+(

*n*-1)

*d*

** nth term or General term (or, last term) of an Arithmetic Progression**:

If ‘

*a*‘ be the first term and ‘

*d*‘ be the common difference then

*a*

_{1}= first term =

*a*=

*a*+(1-1)

*d*

*a*

_{2}= 2nd term =

*a*+

*d*=

*a*+(2-1)

*d*

*a*

_{3}= 3rd term =

*a*+2

*d*=

*a*+(3-1))

*d*

*a*

_{4}= 4th term =

*a*+3

*d*=

*a*+(4-1)

*d*

…

*a*=

_{n}*n*th term =

*a*+(

*n*-1)

*d*

*a _{n}*=

*a*+(

*n*-1)

*d*

in which

*a*= 1st term,

*d*= common difference,

*n*= number of terms

Example 1: **Calculating a Term in an Arithmetic Sequence, Given Two Terms**

Two terms in an arithmetic sequence are *t*_{3}=4 and *t*_{8}=34. What is *t*_{1}?

solution:

*t*_{3}=4 and *t*_{8}=34.

Sketch a diagram. Let the common difference be *d*.

From the diagram,

*t*

_{8}=

*t*

_{3}+5

*d*

Substitute:

*t*

_{8}=34,

*t*

_{3}=4

*d*

Solve for

*d*.

*d*

*d*=6

Then,

*t*

_{1}=

*t*

_{3}-2

*d*

Substitute:

*t*

_{3}=4,

*d*=6

*t*

_{1}=4-2(6)

*t*

_{1}=4-12

*t*

_{1}=-8

Example 2: **Finding a term in an arithmetic sequence**

(A) If the first and te*n*th terms of an arithmetic sequence are 3 and 30, respectively, find the fiftieth term of the sequence.

Solution:

(A) First use Theorem 1 with *a*_{1}=3 and *a*_{10}=30 to find *d*:

*a*=

_{n}*a*

_{1}+(

*n*-1)

*d*

*a*

_{10}=

*a*

_{1}+(10-1)

*d*

30=3+9

*d*

*d*=3

Now find

*a*

_{50}

*a*

_{50}=

*a*

_{1}+(50-1)•3

3+49•3=150

Example 3. Find the 16th and *n*th terms in an arithmetic sequence with the fourth term 15 and eighth term 37.

Steps:

(1) Write the formula for the *n*th term of the arithmetic sequence.

*t*=

_{n}*a*+(

*n*-1)×

*d*

(2) Substitute

*n*=4 and

*t*

_{4}=15 into the formula.

*t*

_{4}=

*a*+3

*d*=15

(3) Substitute

*n*=8 and

*t*

_{8}=37 into the formula

*t*

_{8}=

*a*+7

*d*=37

(4) Solve the simultaneous equations: subtract

*t*

_{4}from

*t*

_{8}to eliminate

*a*

*t*

_{8}–

*t*

_{4})…

*a*+7

*d*-a-3

*d*=37-15

4

*d*=22→

*d*=5½

(5) Substitute

*d*=5½ into

*t*

_{4}and solve for

*a*.

*t*

_{4})…

*a*+3×5½=15

*a*=-1½

(6) To find the

*n*th term of the arithmetic sequence, substitute the values of

*a*and

*d*into the general formula and simplify.

*t*=-1½+(

_{n}*n*-1)×5½

*t*=-1½-5½+5½

_{n}*n*

*t*=5½

_{n}*n*-7

(7) To find the 16th term, substitute

*n*=16 into the formula established in the previous step and evaluate.

*t*

_{16}=5½×16-7=88-7

*t*

_{16}=81

Example 4. Find the missing terms in this arithmetic sequence: {41, *a*, 55, *b*, …}.

Steps:

(1) The first three successive terms are 41, a, 55. Write the rule for the middle term of the three successive terms of an arithmetic sequence.

For *x*, *y*, *z*: *y*=½(*x*+*z*).

(2) Identify the variables.

*x*=41;

*y=a*;

*z*=55

(3) Substitute the values of

*x*,

*y*and

*z*into the formula in step 1 and evaluate.

*a*=½(41+55)=48

(4) Find the common difference. (The second term is now known.)

*d*=

*t*

_{2}–

*t*

_{1}=48-41=7

(5) Find the value of

*b*by adding the common difference to the preceding term.

*b*=55+7=62

(6) State your answer.

so

*a*=48,

*b*=62

Example 5. The 4th term of an arithmetic sequence is 80 and the 12th term is 32.

a. What is the common difference?

b. What is the first term of the sequence?

solution:

a. How to Proceed

(1) Use *a _{n}*=

*a*

_{1}+(

*n*-1)

*d*to write two equation in two variables:

(2) Subtract to eliminate

*a*

_{1}:

(3) Solve for

*d*:

b. Substitute in either equation to find

*a*

_{1}.

*a*

_{1}+3(-6)

80=

*a*

_{1}-18

98=

*a*

_{1}

answer:

*a*.

*d*=-6 b.

*a*

_{1}=98

Example 6. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Solution:

Here, *a*_{11}=38 and *a*_{16}=73 to find *a*_{31}.

Given that: *a*_{11}=*a*+(11-1)*d*=38

*a*+10

*d*=38

*a*=38-10

*d*…(1)

and

*a*

_{16}=73 ⇒

*a*+15

*d*=73

Putting the value of a from equation (1), we get 38-10

*d*+15

*d*=73

*d*=35

*d*=7

Putting the value of

*d*in equation (1), we get

*a*=38-10(7)=-32

Therefore,

*a*

_{31}=

*a*+30

*d*=-32+30(7)=178

Hence, the 31st term is 178.

Example 7. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Solution:

Here, *a*_{3}=12 and *a*_{50}=106 to find *a*_{29}.

Given that: *a*_{3}=*a*+(3-1)*d*=12

*a*+2

*d*=12

⇒

*a*=12-2

*d*…(1)

and

*a*

_{50}=106

*a*+49

*d*=106

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

*d*+49

*d*=106

47

*d*=94

*d*=2

Putting the value of

*d*in equation (1), we get

*a*=12-2(2)=8

Therefore,

*a*

_{29}=

*a*+28

*d*=8+28(2)=64

Hence, the 29th term of the AP is 64.

Let’s read post How to Describe every Arithmetic Sequence?

Ex8. The 2nd term of an arithmetic sequence is 13 and 5th term is 25. What is its 17 term?

(a) 69 (b) 73 (c) 77 (d) 81

Solution:

*a*

_{2}=

*a*+

*d*=13 … (i)

*a*

_{5}=

*a*+4

*d*=25 … (ii)

On subtracting (i) from (ii), we get:

*d*=12

*d*=4.

On putting the value of

*d*in (i), we get

*a*+4=13

*a*=9.

Now,

*a*

_{17}=

*a*+16

*d*=9+16⋅4=73.

Hence, the 17th term is 73.

Answer: (b) 73

Ex9. If 4 times the 4th term of an arithmetic sequence is equal to 18 times its 18th term then find its 22nd term.

Solution:

Let *a* be the first term and *d* be the common difference of the AP. Given 4⋅*a*_{4}=18⋅*a*_{18}.

*a*+3

*d*)=18(

*a*+17

*d*) … [

*a*=

_{n}*a*+(

*n*-1)

*d*]

2(

*a*+3

*d*)=9(

*a*+17

*d*)

2

*a*+6

*d*=9

*a*+153

*d*

7

*a*=-147

*d*

*a*=-21

*d*

*a*+21

*d*=0

*a*_{22}=*a*+(22-1)*d*

*a*_{22}=*a*+2

*d*

*a*

_{22}=0

Hence, the 22nd term of the arithmetic sequence is 0.

Ex10. If 10 times the 10th term of an arithmetic sequence is equal to 15 times the 15th term, show that its 25th term is zero.

Solution:

Let *a* be the first term and *d* be the common difference of the AP. Given 10⋅*a*_{10}=15⋅*a*_{15}.

*a*+9

*d*)=15(

*a*+14

*d*) …[

*a*=

_{n}*a*+(

*n*-1)

*d*]

2(

*a*+9

*d*)=3(

*a*+14

*d*)

2

*a*+18

*d*=3

*a*+42d

*a*=-24d

*a*+24

*d*=0

*a*

_{25}=

*a*+(25-1)

*d*

*a*

_{25}=

*a*+24

*d*

*a*

_{25}=0

Hence, the 25th term of the arithmetic sequence is 0.

Ex11. The 7th term of an arithmetic sequence is -1 and its 16th term is 17. The *n*th term of the arithmetic sequence is

(a) (3*n*+8) (b) (4*n*-7) (c) (15-2*n*) (d) (2*n*-15)

Solution:

Let *a* be the first term and *d* be the common difference of the AP. Then,

*n*th term of the AP, *a _{n}*=

*a*+(

*n*-1)

*d*

Given

*a*

_{7}=-1.

*a*+6

*d*=-1 … (1)

Given

*a*

_{16}=17.

*a*+15

*d*=17 … (2)

Subtracting (1) from (2), we get (

*a*+15

*d*)-(

*a*+6

*d*)=17-(-1)

*d*=18

*d*=2.

Putting

*d*=2 in (1), we get

*a*+6⋅2=-1

*a*=-1-12=-13

∴

*n*th term of the AP,

*a*=-13+(

_{n}*n*-1)⋅2=2

*n*-15

Answer: (d) (2

*n*-15)

Ex12. The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term. Solution:

Given, *a*_{6}=19, *a*_{17}=41 then

*a*

_{6}=

*a*+(6-1)

*d*

19=

*a*+5

*d*… (1)

*a*

_{17}=

*a*+(17-1)

*d*

41=

*a*+16

*d*… (2).

Subtract (1) from (2).

Substitute

*d*=2 in (l)

19=

*a*+5⋅2

9=

*a*

∴ 40th term

*a*

_{40}=

*a*+(40-1)⋅

*d*

*a*

_{40}=9+39⋅2

=9+78

=87

Ex13. The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find 26th term.

Solution:

Given,

*a*

_{10}=41,

*a*

_{18}=73,

*a*=

_{n}*a*+(

*n*-1)⋅

*d*

⇒

*a*

_{10}=

*a*+(10-1)⋅

*d*

41=

*a*+9d …(1)

*a*

_{18}=

*a*+(18-1)

*d*

73=

*a*+17

*d*…(2)

Subtract (1) from (2).

Substitute

*d*=4 in (1).

*a*+9⋅4=41

*a*=41-36

*a*=5

26th term

*a*

_{26}=

*a*+(26-1)

*d*

=5+100

=105

∴ 26th term

*a*

_{26}=105.