π Using particular sum formulae as follows:

(i). *S _{n}*=Β½

*n*(

*a*+

*β*)

(ii).

*S*=Β½

_{n}*n*(

*a*+

*a*)

_{n}(iii).

*S*=Β½

_{n}*n*(2

*a*+(

*n*-1)

*d*)

(iv).

*S*=

_{n}*S*

_{n-1}+

*a*, i.e.,

_{n}*a*=

_{n}*S*–

_{n}*S*

_{n-1}

π Example 1. In an arithmetic sequence, the first term is 2, the last term is 29 and the sum of all the terms is 155. Find the common difference.

β Solution:

Here, *a*=2, *β*=29 and *S _{n}*=155.

Let

*d*be the common difference of the given arithmetic sequence and

*n*be the total number of terms. Then,

*a*=29.

_{n}*a*+(

*n*-1)

*d*=29

2+(

*n*-1)

*d*=29 β¦ (i)

The sum of

*n*terms of an arithmetic sequence is given by

*S*=Β½

_{n}*n*(

*a*+

*β*)=155

Β½

*n*(2+29)=Β½β

*n*β 31=155

*n*=10.

Putting the value of

*n*in (i), we get

*d*=29

9

*d*=27

*d*=3.

Thus, the common difference of the given arithmetic sequence is 3.

π Ex2. In an arithmetic sequence, the first term is -4, the last term is 29 and the sum of all its terms is 150. Find its common difference.

β Solution:

Suppose there are *n* terms in the arithmetic sequence.

Here, *a*=-4, *β*=29 and *S _{n}*=150.

*S*=Β½

_{n}*n*(

*a*+

*β*)

Β½

*n*(-4+29)=150

*n*=(150β 2)/25=12.

Thus, the arithmetic sequence contains 12 terms. Let

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

*a*

_{12}=29

[

*a*=

_{n}*a*+(

*n*-1)

*d*]

-4+(12-1)β

*d*=29

11

*d*=29+4=33

*d*=3.

Hence, the common difference of the arithmetic sequence is 3.

π Ex3. The first and last terms of an arithmetic sequence are 5 and 45 respectively. If the sum of all its terms is 400, find the common difference and the number of terms.

β Solution:

Suppose there are *n* term in the arithmetic sequence.

Here, *a*=5, *β*=45 and *S _{n}*=400.

*S*=400

_{n}*S*=Β½

_{n}*n*(

*a*+

*β*)

Β½

*n*(5+45)=400

*n*β Β½β 50=400

*n*=400Γ·25

*n*=16.

Thus, there are 16 terms in the arithmetic sequence. Let

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

*a*

_{16}=45

[

*a*=

_{n}*a*+(

*n*-1)

*d*]

*d*=45

15

*d*=45-5=40

*d*=40/15=8/3.

Hence, the common difference of the arithmetic sequence is 8/3.

π Ex4. In an arithmetic sequence: given *a*=8, *a _{n}*=62,

*S*=210, find

_{n}*n*and

*d*.

β Solution:

Here,

*a*=8,

*a*=62 and

_{n}*S*=210.

_{n}The sum of

*n*terms of an arithmetic sequence is given by

*S*=Β½

_{n}*n*(

*a*+

*a*)

_{n}210=Β½

*n*β (8+62)

210=35

*n*

6=n

β

*a*=

_{n}*a*+(

*n*-1)

*d*

62=8+(6-1)

*d*

54=5

*d*β

*d*=54/5

π Ex5. In an arithmetic sequence: given *a*=3, *n*=8, *S*=192, find *d*.

β Solution:

Here, *a*=3, *n*=8 and *S*=192. The sum of *n* terms of an arithmetic sequence is given by

*S*=Β½

_{n}*n*(

*a*+

*a*)

_{n}192 =8/2[3+

*a*]

_{n}192=4[3+

*a*]

_{n}3+

*a*=192/4=48

_{n}*a*=45

_{n}β

*a*=

_{n}*a*+(

*n*-1)

*d*

45=3+(8-1)

*d*

42=7

*d*

*d*=6

π Ex6. An arithmetic progression has 3 as its first term. Also, the sum of the first 8 terms is twice the sum of the first 5 terms. Find the common difference.

β Solution:

We are given that *a*=3. We are also given some information about the sums *S*_{8} and *S*_{5}, and we want to find the common difference. So we shall use the formula

*S*=Β½

_{n}*n*(2

*a*+(

*n*-1)

*d*)

for the sum of the first

*n*terms. This tells us that

*S*

_{8}=Β½Γ8Γ(6+7

*d*)

and that

*S*

_{5}=Β½Γ5Γ(6+4

*d*).

So, using the given fact that

*S*

_{8}=2

*S*

_{5}, we see that

*d*)=2ΓΒ½Γ5Γ(6+4

*d*)

4Γ(6+7

*d*)=5Γ(6+4

*d*)

24+28

*d*=30+20

*d*

8

*d*=6

*d*=ΒΎ

π Ex7. In an arithmetic sequence, the first term is 22, *n*th terms is -11 and sum of first *n* terms is 66. Find the number *n* and the common difference.

β Solution:

Here, *a*=22, *a _{n}*=-11 and

*S*=66.

_{n}Let

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

Then,

*a*=-11

_{n}*a*+(

*n*-1)

*d*=-11

=22+(

*n*-1)

*d*=-11

(

*n*-1)

*d*=-33 β¦ (i)

The sum of

*n*terms of an arithmetic sequence is given by

*S*=Β½

_{n}*n*[2

*a*+(

*n*-1)

*d*]=66

[Substituting the value off (

*n*-1)

*d*from (i)]

Β½

*n*[2β 22+(-33)]=Β½

*n*β 11=66

*n*=12.

Putting the value of

*n*in (i), we get:

*d*=-33

*d*=-3.

Thus,

*n*=12 and

*d*=-3.

π Ex8. If the sum of first *n* terms is (3*n*^{2}+5*n*), find its common difference.

β Solution:

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

*n*terms of the arithmetic sequence.

*S*=3

_{n}*n*

^{2}+5

*n*

*S*

_{n-1}=3(

*n*-1)

^{2}+5(

*n*-1)

=3(

*n*

^{2}-2

*n*+1)+5(

*n*-1)

=3

*n*

^{2}–

*n*-2.

Now,

*n*th term of arithmetic sequence,

*a*=

_{n}*S*–

_{n}*S*

_{n-1}

*n*

^{2}+5

*n*)-(3

*n*

^{2}–

*n*-2)

=6

*n*+2.

Let

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

*d*=

*a*–

_{n}*a*

_{(n-1)}

=(6

*n*+2)-[6(

*n*-1)+2]

=6

*n*+2-6(

*n*-1)-2

=6.

Hence, the common difference of the arithmetic sequence is 6.

π Ex9. The sum of first *n* terms of an arithmetic sequence is (3*n*^{2}+6*n*). The common difference of the arithmetic sequence is

β Solution:

Let

*S*denotes the sum of first

_{n}*n*terms of the arithmetic sequence.

*S*=3

_{n}*n*

^{2}+6

*n*

β

*S*

_{n-1}=3(

*n*-1)

^{2}+6(

*n*-1)

=3(

*n*

^{2}-2

*n*+1)+6(

*n*-1)

=3

*n*

^{2}-3.

So,

*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.

Let

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

*d*=

*a*–

_{n}*a*

_{(n-1)}=(6

*n*+3)-[6(

*n*-1)+3]

=6

*n*+3-6(

*n*-1)-3

=6.

Thus, the common difference of the arithmetic sequence is 6.

Answer: (a) 6