# Series and Sequence [Diploma,Aptitude test,SEE]

### Series and Sequence

A sequence is a set of numbers written in a particular order.
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
Sₙ = u₁+u₂+u₃+........+uₙ

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 (tₙ) =a+(n-1)d
Hence,tₙ) =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, a=2,
number of terms, n=45,
Common difference, d=3
So, The 45th term is:
t45=a+(n1)d
=2+(451)3
=2+44×3
=134

2.Find the 20th term of the arithmetic sequence 5,8,11,54,... .
Solution:
Given Sequence,
5,8,11,54,... .
a=5,d=85=3
So, t20 =a+(n1)d
=5+(201)(3)
=5+19×3
=62

3.Find 18th term in the sequence 7, 13, 19, 25,...
Solution:
Given Sequence,

7, 13, 19, 25, ...
Here, a = 7, d = 13 – 7 = 6
18th term, t18=a+(n1)d
=7+(181)6
=7+102

=109

4.Find the 46th term of the arithmetic sequence 2,5,8,11,⋯ .
Solution:
Given sequence,
2,5,8,11,⋯ .
Here, first term, a=2,
number of terms, n=46,
Common difference, d=3
So, The 46th term is:
t40=a+(n1)d
=2+(461)3
=2+45×3
=137

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, l = 72
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 (tₙ) =64
i.e.  a+ (n-1) d =64
or, 2+ (n-1) 2 =64
or, 2+2n-2 =64
∴n= 32

7. The 6th term of  an A.P. is 24 and the 8tterm is 30. Are 63 and 65 terms of this A.P.?
Solution:
Here, 6th term =24
i.e. a+5d = 24 ...(i)
Again,8th term = 30 ...(i)
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 nth term of A.P.
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 19th term of A.P.

If possible, let 65 is the pth term of A.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 nth term of the series 5+3+1+ ... is same as the nth 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
nth term = a+(n-1)d
=5+(n-1)(-2)=7-2n
For the series 9+6+3+ ...
First term (a) =9
Common difference (d) = 6-9 =-3
nth term  =  a+(n-1)d
=9+(n-1) (-3) = 12-3n
∴n=4