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__Series and Sequence__

A sequence is a set of numbers written in a particular order.__Series and Sequence__

Eg.2,4,6,8,10...... is a set of even numbers.

2,3,5,7,11....... is a set of prime numbers.

__Series__A series is a sum of the terms of a sequence. If there are n terms in the sequence and we evaluate the sum then we often write

Sâ‚™ = u₁+u₂+u₃+........+uâ‚™

__Arithmetic progression__Lets consider the two common sequences:

3,2,1,0...... and 2,4,6,8....... .

These sequence start with a particular first term and then to get successive terms. we just add or subtract instead because that is just the same as adding a negative constant. e.g. in a sequence 3,2,1,0...... and 2,4,6,8,10 the difference between the consecutive terms in each sequence is constant. In the first sequence, we get -1 when we subtract t₂- t₁= t₄ -t₃=-1 and we add 2 to get next term in the second sequence.

If a stands for first term of the sequence and d stands for the common difference.

First term (t₁)= a+(n-1)d = a+(1-1)d= a

Second term (t₂)= a+(n-1)d =a+(2-1)d =a+d

Third term(t₃)= a+(n-1)d =a+(3-1)d= a+2d

Fourth term(t₄)= a+(n-1)d = a+(4-1)d=a+3d

nth term (tâ‚™) =a+(n-1)d

Hence,tâ‚™) =a+(n-1)d

Note: [sometimes,we also write l for the last term of a finite sequence.so,we also write l=a+(n-1)d]

**Practice Set 3****1.Find the 45th term of the arithmetic sequence 2,5,8,11,⋯ .**

**Solution:**

Given sequence,

2,5,8,11,⋯ .

Here, first term,

number of terms,

Common difference,

So, The

2. Find the 20 t h term of the arithmetic sequence 5,8,11,54,... .

**Solution:**

Given Sequence,

5,8,11,54,... .

So,

**3.Find**18 t h term in the sequence 7, 13, 19, 25,...

**Solution:**

Here, a = 7, d = 13 – 7 = 6

**4.Find the**46 t h term of the arithmetic sequence 2,5,8,11,⋯ .

**Solution:**

Given sequence,

2,5,8,11,⋯ .

Here, first term,

number of terms,

Common difference,

So, The

5. Find the number of terms in the series 8, 12, 16, . . .,72.

**Solution:**

Given Sequence,

8, 12, 16, . . .,72.

First term, a = 8

Last term,

Common difference, d = 12 – 8 = 4.

Number of terms:?

We have,last term (l)= a+(n-1)d

or, 72 =8+(n-1) 4

or, 72 =8+ 4n- 4

or, 68=4n

n= 17

**6. How many terms are there in the series 2+4+6+ ... +64?**

**Solution:**

First term (a) = 2

Common difference (d) =4-2=2

Here, Last term (tâ‚™) =64

i.e. a+ (n-1) d =64

or, 2+ (n-1) 2 =64

or, 2+2n-2 =64

∴n= 32

**7. The**6 t h term of an A.P. is 24 and the 8t h term is 30. Are 63 and 65 terms of this A.P.?

**Solution:**

Here,

i.e. a+5d = 24 ...(i)

Again,8

Subtracting (i) from (ii) we get,

2d = 6

d= 3

Putting the value of d in equation (i) we get,

a+5d=24

or,a+5×3 = 24

or,a = 9

If possible, let 63 is the n

Then, a+(n-1)d= 63

or,9+(n-1)3 = 63

or,9+3n-3 = 63

or,3n = 57

or,n=19

∴63 is the 19

If possible, let 65 is the p

Then, a+(p-1)d = 65

or,9+(p-1)3 = 65

or,3p+6 = 65

or, p=59/3

∴65 is not any term of the A.P. because p is fraction, which is not possible.

**8. The n**t h term of the series 5+3+1+ ... is same as the nt h term of the series 9+6+3+ ...

**Find n.**

**Solution:**

For the series 5+3+1+ ...

First term (a) =5

Common difference (d) =3-5=-2

n

=5+(n-1)(-2)=7-2n

For the series 9+6+3+ ...

First term (a) =9

Common difference (d) = 6-9 =-3

n

=9+(n-1) (-3) = 12-3n

∴n=4