Problem 1. If the sum of the first *n* terms of an arithmetic sequence is given by *S _{n}*=2

*n*

^{2}+5

*n*, Find the

*n*th term of the arithmetic sequence

Solution: We shall use the formula

*a*=

_{n}*S*–

_{n}*S*

_{(n-1)};

*n*=2,3,4, … ;

*a*

_{1}=

*S*

_{1}

Where

*S*

_{0}=0

*a*=2

_{n}*n*

^{2}+5

*n*-[2(

*n*-1)

^{2}+5(

*n*-1)]

=2

*n*

^{2}+5

*n*-[2(

*n*

^{2}-2

*n*+1)+5

*n*-5]

=2

*n*

^{2}+5

*n*-2

*n*

^{2}+4

*n*-2-5

*n*+5

*a*=4

_{n}*n*+3

Thus,

*n*th term of the arithmetic sequence is 4

*n*+3.

P2. The sum of first *n* terms of an arithmetic sequence is (5*n*–*n*^{2}) The *n*th term of the arithmetic sequence is

(a) (5-2*n*) (b) (6-2*n*) (c) (2*n*-5) (d) (2*n*-6)

Sol:

Let *S _{n}* denotes the sum of first

*n*terms of the arithmetic sequence.

*S*=5

_{n}*n*–

*n*

^{2}

*S*

_{(n-1)}=5(

*n*-1)-⋅(

*n*-1)

^{2}=5

*n*-5-

*n*

^{2}+2

*n*-1

=7

*n*–

*n*

^{2}-6

∴

*n*th term of the arithmetic sequence,

*a*=

_{n}*S*–

_{n}*S*

_{(n-1)}

*n*–

*n*

^{2})-(7

*n*–

*n*

^{2}-6)

=6-2

*n*

Thus, the

*n*th term of the arithmetic sequence is (6-2

*n*).

Answer: (b) (6-2

*n*)

P3. The sum of the first it terms of an arithmetic sequence is (4*n*^{2}+2*n*). The *n*th term of this arithmetic sequence is

(a) (6*n*-2) (b) (7*n*-3) (c) (8*n*-2) (d) (8*n*+2)

Solution:

Let *S _{n}* denotes the sum of first

*n*terms of the arithmetic sequence.

*S*=4

_{n}*n*

^{2}+2

*n*

⇒

*S*

_{(n-1)}=4⋅(

*n*-1)

^{2}+2(

*n*-1)

=4(

*n*

^{2}-2

*n*+1)+2(

*n*-1)

=4

*n*

^{2}-6

*n*+2

∴

*n*th term of the arithmetic sequence,

*a*=

_{n}*S*–

_{n}*S*

_{(n-1)}

*n*

^{2}+2

*n*)-(4

*n*

^{2}-6

*n*+2)

=8

*n*-2

Thus, the

*n*th term of thee arithmetic sequence is (8

*n*-2)

Answer: (c) (8

*n*-2)

P4. If the sum of *n* terms of an arithmetic sequence is 2*n*+3*n*^{2}. Find the *n*th term.

Solution:

We have *S _{n}*=2

*n*+3

*n*

^{2}

*S*

_{(n-1)}=2(

*n*-1)+3(

*n*-1)

^{2}

=2(

*n*-1)+3(

*n*

^{2}-2

*n*+1)

=2

*n*-2+3

*n*

^{2}-6

*n*+3

*S*

_{(n-1)}=3

*n*

^{2}-4

*n*+1

*n*th term =

*a*=

_{n}*S*–

_{n}*S*

_{(n-1)}

*n*+3

*n*

^{2}-(3

*n*

^{2}-4

*n*+1)

=2

*n*+3

*n*

^{2}-3

*n*

^{2}+4

*n*-1

*a*=6

_{n}*n*-1

P5. The sum of the first *n* terms of an arithmetic sequence is (3*n*^{2}+6*n*). Find the *n*th term and the 15th term of this arithmetic sequence.

Solution:

Let *S _{n}* denotes the sum of first

*n*terms of the arithmetic sequence.

*S*=3

_{n}*n*

^{2}+6n

*S*

_{(n-1)}=3(

*n*-1)

^{2}+6(

*n*-1)

=3(

*n*

^{2}-2

*n*+1)+6(

*n*-1)

=3

*n*

^{2}-3

∴

*n*th term of the arithmetic sequence,

*a*.

_{n}*S*–

_{n}*S*

_{(n-1)}

=(3

*n*

^{2}+6

*n*)-(3

*n*

^{2}-3)

=6

*n*+3

Putting

*n*=15, we get

*a*

_{15}=6⋅15+3=90+3=93

Hence, the

*n*th term is (6

*n*+3) and 15th term is 93.

P6. The sum of the first *n* terms of an arithmetic sequence is given by *S _{n}*=(3

*n*

^{2}–

*n*). Find its

(i)

*n*th term, (ii) first term and (iii) common difference.

Solution:

Given

*S*=(3

_{n}*n*

^{2}–

*n*) … (a)

Replacing

*n*by (

*n*-1) in (a), we get:

*S*

_{(n-1)}=3(

*n*-1)

^{2}-(

*n*-1)

=3(

*n*

^{2}-2

*n*+1)-

*n*+1

=3

*n*

^{2}-7

*n*+4

(i) Now,

*a*=

_{n}*S*–

_{n}*S*

_{(n-1)}

=(3

*n*

^{2}–

*n*)-(3

*n*

^{2}-7

*n*+4)=6

*n*-4

∴

*n*th term,

*a*=(6

_{n}*n*-4) … (b)

(ii) Putting

*n*=1 in (b), we get:

*a*

_{1}=(6⋅1)-4=2

(iii) Putting

*n*=2 in (b), we get:

*a*

_{2}=(6⋅2)-4=8

∴ Common difference,

*d*=

*a*

_{2}–

*a*

_{1}=8-2=6

P7. If the sum of first in terms of an arithmetic sequence is (2*m*^{2}+3*m*) then what is its second term? Solution:

Let *S _{m}* denotes the sum of first In terms of the arithmetic sequence.

*S*=2

_{m}*m*

^{2}+3

*m*

=2⋅(

*m*-1)

^{2}+3(

*m*-1)=2(

*m*

^{2}-2

*m*+1)+3(

*m*-1)=2

*m*

^{2}–

*m*-1

Now,

mth term of arithmetic sequence,

*a*=

_{m}*S*–

_{m}*S*

_{(m-1)}

*a*=2

_{m}*m*

^{2}+3

*m*-(2

*m*

^{2}–

*m*-1)

*a*=4

_{m}*m*+1

Putting

*m*=2, we get

*a*

_{2}=4⋅2+1=9

Hence, the second term of the arithmetic sequence is 9.

P8. The sum of *n* terms is given by *S _{n}*=½

*n*(1+

*n*). Determine

*a*

_{5}.

Solution:

*S*

_{5}=½⋅5⋅(1+5)=15

*S*

_{4}=½⋅4⋅(1+4)=10

*a*

_{5}=15-10=5

P9. (Prizes) A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by $100. Find the amounts of the first and last prize.

Solution:

Given *n*=10, *d*=100 and *S*_{10}=8500.

Find the value of *a*_{1}.

*S*=½

_{n}*n*[2

*a*

_{1}+(

*n*-1)

*d*]

*S*

_{10}=½⋅10⋅[2

*a*

_{1}+(10-1)100]

8500=5⋅[2

*a*

_{1}+900]

*a*

_{1}=400

Find the value of

*a*

_{10}.

*a*

_{10}=

*a*

_{1}+(

*n*-1)

*d*

=400+(10-1)100

=1300

Answer: $400 and $1300

P10. In an AP: given *a*_{3}=15, *S*_{10}=125, find *d* and *a*_{10}.

Solution:

Here, *a*_{3}=15 and *S*_{10}=125.

*a*

_{3}=15

*a*+(3-1)

*d*=15

*a*+2

*d*=15

*a*=15-2

*d*… (1)

The sum of

*n*terms of an arithmetic sequence is given by

*S*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*]

*S*

_{10}=½⋅10⋅[2

*a*+(10-1)

*d*]

125=5[2

*a*+9

*d*]

2

*a*+9

*d*=25

Putting the value of

*a*from equation (1), we get

*d*)+9

*d*=25

30-4

*d*+9

*d*=25

5

*d*=-5

*d*=-1

Putting the value of

*d*in equation (1), we get

*a*=15-2(-1)=17

*a*=

_{n}*a*+(

*n*-1)

*d*

*a*

_{10}=17+(10-1)(-1)= 17-9 =8

*a*

_{10}=8

P11. In an AP: given *d*=5, *S*_{9}=75, find *a* and *a*_{9}.

Solution:

Here, *d*=5 and *S*_{9}=75. The sum of *n* terms of an arithmetic sequence is given by

*S*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*]

*S*

_{9}=½⋅9⋅[2

*a*+(9-1)(5)]

75=½⋅9⋅[2

*a*+40]

75=9

*a*+180

75-180=9a

*a*=-105/9=-35/3=-11⅔

⇒

*a*=

_{n}*a*+(

*n*-1)

*d*

*a*

_{9}=-11⅔+(9-1)5

*a*

_{9}=-11-⅔+40

*a*

_{9}=40-11-⅔

*a*

_{9}=28⅓

P12. If the sum of a certain number of terms of the arithmetic sequence 25, 22, 19, is 116. Find the last term

Solution:

Let the sum of *n* terms of the given arithmetic sequence be 116.

*S*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*]

Here,

*a*=25 and

*d*=22-25=-3

*n*[2⋅25+(

*n*-1)(-3)]

232=

*n*⋅[50-3

*n*+3]

232=53

*n*-3

*n*

^{2}

3

*n*

^{2}-53

*n*-232=0

3

*n*

^{2}-24

*n*-29

*n*+232=0

3n(

*n*-8)-29(

*n*-8)=0

(

*n*-8)(3

*n*-29)=0

*n*=8 or

*n*=29/3

However,

*n*cannot be equal to 29/3. Therefore,

*n*=8. ∴ The last term

*a*

_{8}=

*a*

_{1}+

*d*(8-1)

*a*

_{8}=25-3⋅7=4

Thus, the last term of the arithmetic sequence is 4.

P13. The sum of the first eight terms of an arithmetic sequence {*a _{n}*} is 24; the sixth term is 0. Find a formula for

*a*.

_{n}Solution:

For

*a*, first find

_{n}*a*and

*d*. Since:

*a*

_{6}=

*a*+5

*d*,

*a*+5

*d*=0. Express

*S*

_{8}in terms of

*a*and

*d*,

*S*

_{8}=½⋅8⋅[

*a*+(

*a*+7

*d*)]=4(2

*a*+7

*d*)

Since we are given

*S*

_{8}=24, Equation (3) states that 4(2

*a*+7

*d*)=24. This gives a pair of equations to solve for

*a*and

*d*.

We find

*d*=-2 and

*a*=10. Therefore, the

*n*th term is

*a*=

_{n}*a*+(

*n*-1)

*d*=10+(

*n*-1)(-2)=12-2

*n*.

P14. In an arithmetic sequence, the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is -112.

Solution:

First term =2

Let *d* be the common difference of the arithmetic sequence

Therefore, the arithmetic sequence is 2, 2+*d*, 2+2*d*, 2+3*d*,

Sum of first five terms =10+10*d*

Sum of next five terms =10+35*d*

According to the given condition,

*d*=¼(10+35

*d*)

40+40

*d*=10+35

*d*

30=-5

*d*

*d*=-6

*a*

_{20}=

*a*+(20-1)

*d*=2+(19)(-6)=2-114=-112

Thus, the 20th term of the arithmetic sequence is -112.

P15. The sum of the first 14 terms of and arithmetic sequence is 1505 and its first term is 10. Find its 25th term.

Solution:

Let *d* be the common difference of the arithmetic sequence.

Here, *a*=10 and *n*=14.

Given *S*_{14}=1505.

*S*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*]

½⋅14⋅[2⋅10+(14-1)⋅

*d*]=1505

7(20+13

*d*)=1505

20+13

*d*=215

13

*d*=215-20=195

*d*=15

∴ 25th term of the arithmetic sequence,

*a*

_{25}=10+(25 -1)⋅15

=10+360

=370.

Hence, the required term is 370.