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
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
Solution:
Given Sequence,
5,8,11,54,... .
So,
Here, a = 7, d = 13 – 7 = 6
4.Find the
Solution:
Given sequence,
2,5,8,11,⋯ .
Here, first term,
number of terms,
Common difference,
So, The
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
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
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