Example 1: **Find the First Three Terms**

Find the first three terms of an arithmetic series in which *a*_{1}=9, *a _{n}*=105, and

*S*=741.

_{n}Solution steps:

Step 1. Since you know

*a*

_{1},

*a*, and

_{n}*S*, use

_{n}*S*=½

_{n}*n*(

*a*

_{1}+

*a*) to find n.

_{n}*S*=½

_{n}*n*(

*a*

_{1}+

*a*)

_{n}741=½

*n*(9+105)

741=57

*n*

13

*n*

Step 2. Find

*d*.

*a*=

_{n}*a*

_{1}+(

*n*-1)

*d*

105=9+(13-1)

*d*

96=12d

8=d

Step 3. Use

*d*to determine 112 and 113.

The first three terms are 9, 17, and 25.

Ex2: **Determining the First Few Terms Given the Sum, Common Difference, and One Term**

An arithmetic series has *S*_{20}=143⅓, *d*=⅓ and *t*_{20}=10⅓; determine the first 3 terms of the series.

solution:

*S*_{20} and *t*_{20} are known, so use this rule to determine *t*_{1}:

*S*=½

_{n}*n*(

*t*

_{1}+

*t*)

_{n}Substitute:

*n*=20,

*S*

_{20}=143⅓,

*t*

_{20}=10⅓

*t*

_{1}+10⅓)

Simplify.

*t*

_{1}+10+⅓)

143⅓=10

*t*

_{1}+100+3⅓

143⅓-3⅓-100=10

*t*

_{1}

Solve for

*t*

_{1}.

*t*

_{1}

4=

*t*

_{1}

The first term is 4 and the common dlfference is ⅓.

So, the first 3 terms of the series are written as the partial sum:

Let’s open a particular post in new tab Sum Formula for every Arithmetic Series with Given the First and Last Terms.

Ex3. The sum of three numbers in an A.P. is 12 and the sum of their cubes is 408. Find them.

Solution:

Let the required numbers be

*a*–

*d*,

*a*,

*a*+

*d*

According to 1st condition:

*a*–

*d*)+

*a*+(

*a*+

*d*)=12

*a*–

*d*+

*a*+

*a*+

*d*=12

3

*a*=12

*a*=4

According to 2nd given condition:

*a*–

*d*)

^{3}+

*a*

^{3}+(

*a*+

*d*)

^{3}=408

*a*

^{3}–

*d*

^{3}-3

*a*

^{2}

*d*+

*a*

*d*

^{2}+

*a*

^{3}+

*a*

^{3}+

*d*

^{3}+

*d*

^{3}+3

*a*

^{2}

*d*+3

*a*

*d*

^{2}=408

3

*a*

^{3}+6

*a*

*d*

^{2}=408

3(4)

^{3}+6(4)

*d*

^{2}=408

24

*d*

^{2}=408-192

*d*

^{2}=9

*d*=±3

When

*a*=4,

*d*=3 then number are

*a*–

*d*,

*a*,

*a*+

*d*

i.e. 4-3, 4 , 4+3 i.e. 1, 4, 7

when

*a*=4,

*d*=-3 then numbers are

*a*–

*d*,

*a*,

*a*+

*d*

4+3, 4, 4-3 i.e., 7, 4, 1

Hence the required numbers are 1, 4, 7 or 7, 4, 1

__Note__:

The problem containing three or more numbers in A.P. whose sum in given it is often to assume the number as follows.

If the required numbers in A.P are odd i.e. 3, 5, 7 etc. Then take ‘*a*’ (first term) as the middle number and *d* as the common difference.

Thus three numbers are *a*–*d*, *a*, *a*+*d*. If the required numbers in A.P are even i.e. 2, 4, 6, etc. then take *a*–*d*, *a*+*d* as the middle numbers and 2d as the common difference.

Thus four numbers are *a*-3*d*, *a*–*d*, *a*+*d*, *a*+3*d* and six numbers are:

*a*-5*d*, *a*-3*d*, *a*–*d*, *a*+*d*, *a*+3*d*, *a*+5*d* etc.

Ex4. In an AP. It is given that *S*_{5}+*S*_{7}=167 and *S*_{10}=235, then find the AP, where *S _{n}* denotes the sum of its first

*n*terms.

Solution:

If you meet trouble to understand this solution, let’s open a particular post in new tab How to Form the Sum of the First n Terms of every Arithmetic Series?

Let

*a*be the first term and

*d*be the common difference of thee AP. Then,

*S*

_{5}+

*S*

_{7}=167

*S*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*]

½⋅5⋅(2

*a*+4

*d*)+½⋅7⋅(2

*a*+6

*d*)=167

5

*a*+10

*d*+7

*a*+21

*d*=167

12

*a*+31

*d*=167… (1)

Also,

*S*

_{10}=235

½⋅10⋅(2

*a*+9

*d*)=235

5(2

*a*+9

*d*)=235

2

*a*+9

*d*=47

Multiplying both sides by 6, we get

*a*+54

*d*=282… (2)

Subtracting (1) from (2), we get

*a*+54d-12

*a*-31

*d*=282-167

23

*d*=115

*d*=5

Putting

*d*=5 in (1), we get

*a*+31⋅5=167

12

*a*+155=167

12

*a*=167-155=12

*a*=1.

Hence, the AP is 1, 6, 11, 16, … .

Ex5. The sum of the first 7 terms of an AP is 182. If its 4th and 17th terms are in the ratio 1∶5, find the AP.

Solution:

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

*S*

_{7}=182

*S*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*]

½⋅7⋅(2

*a*+6

*d*)=182

*a*+3

*d*=26 … (1)

Given

*a*

_{4}=

*a*

_{17}=1∶5.

*a*=

_{n}*a*+(

*n*-1)

*d*]

(

*a*+3

*d*)/(

*a*+16

*d*)=⅕

5

*a*+15

*d*=

*a*+16d

*d*=4

*a*… (2)

Solving (1) and (2), we get

*a*+3⋅4

*a*=26

13

*a*=26

*a*=2.

Putting

*a*=2 in (2), we get

*d*=4⋅2=8.

Hence, the required AP is 2, 10, 18, 26, … .

Ex6. The sum first 10 terms of an AP is -150 and the sum of its next 10 terms is -550. Find the AP.

Solution:

Let *a* be the first term and *d* be the common difference of the AP. Then, *S*_{10}=-150 (Given).

*S*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*]

½⋅10⋅(2

*a*+9

*d*)=-150

5(2

*a*+9

*d*)=-150

2

*a*+9

*d*=-30 … (1)

It is given that the sum of its next 10 terms is -550.

Now,

*S*

_{20}= Sum of first 20 terms = Sum of first 10 terms + Sum of the next 10 terms =

-150+(-550)=-700.

*S*

_{20}=-700

½⋅20⋅(2

*a*+19

*d*)=-700

10(2

*a*+19

*d*)=-700

2

*a*+19

*d*=-70… (2)

Subtracting (1) from (2), we get

*a*+19

*d*)-(2

*a*+9

*d*)=-70-(-30)

10

*d*=-40

*d*=-4

Putting

*d*=-4 in (1), we get

*a*+9⋅(-4)=-30

2

*a*=-30+36=6

*a*=3

Hence, the required AP is 3, -1 -5, -9, … .

Ex7. A sum of $700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is $20 less than its preceding prize, find the value of each prize.

Solution:

Let the value of the first prize be a.

Since the value of each prize is 20 less than its preceding prize, so the values of the prizes are in AP with common difference $20.

*a*+(40-1)

*d*]=36000

20(2

*a*+39

*d*)=36000

2

*a*+39

*d*=1800… (2)

Number of cash prizes to be given to the students,

*n*=7.

Total sum of the prizes,

*S*

_{7}=$700

Using the formula,

*S*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*], we get

*S*

_{7}=½⋅7⋅[2

*a*+(7-1)⋅(-20)]=700

½⋅7⋅(2

*a*-120)=700

7

*a*-420=700

7

*a*=700+420=1120

*a*=160

Thus, the value of the first prize is $160.

Hence, the value of each prize is $160, $140, $120, $100, $80, $60 and $40.

**Find the first three terms of each arithmetic series Ex8 to Ex9.**

Ex8. *a*_{1}=8, *a _{n}*=100,

*S*=1296

_{n}Solution:

*S*=½

_{n}*n*⋅(

*a*

_{1}+

*a*)

_{n}1296

*n*⋅½⋅(8+100)

1296

*n*⋅54

*n*=1296÷54=24

*a*=

_{n}*a*

_{1}+(

*n*-1)

*d*

100=8+(24-1)

*d*

23

*d*=92

*d*=4

Therefore, the first three terms are 8, (8+4) or 12, (12+4) or 16.

Answer: 8, 12, 16

Ex9. *n*=18, *a _{n}*=112,

*S*=1098

_{n}Solution:

*S*=½

_{n}*n*⋅(

*a*

_{1}+

*a*)

_{n}1098=½⋅18⋅(

*a*

_{1}+112)

*a*

_{1}+112=1098÷9

*a*

_{1}=122-112

*a*

_{1}=10

*a*=

_{n}*a*

_{1}+(

*n*-1)

*d*

112=10+(18-1)

*d*

17

*d*=102

*d*=6

Therefore, the first three terms are 10, (10+6) or 16, (16+6) or 22.

Answer: 10, 16, 22