# Get Terms from Sequences

Describing sequences
Sequences of numbers play an important part in our everyday life. For example, the following sequence:

2.25, 2.37, 2.58, 2.57, 2.63, …

gives the end-of-day trading price (for 5 consecutive days) of a share in an electronics company. It looks like the price is on the rise, but is it possible to accurately predict the future price per share of the company?
The following sequence is more predictable:

10000, 9000, 8100, ….

This is the estimated number of radioactive decays of a medical compound each minute after administration to a patient. The compound is used to diagnose tumours. In the first minute, 10 000 radioactive decays are predicted; during the second minute, 9000, and so on. Can you predict the next number in the sequence? You’re correct if you said 7290. Each successive term here is 90% of, or 0.90 times, the previous term.

Sequences are strings of numbers. They can be finite in number or infinite. Number sequences may follow an easily recognisable pattern or they may not. A great deal of recent mathematical work has gone into deciding whether certain strings follow a pattern (in which case subsequent terms could be predicted) or whether they are random (in which case subsequent terms cannot be predicted). This work forms the basis of chaos theory, speech recognition software for computers, weather prediction and stock market forecasting, to name but a few uses. The list is almost endless. The image on the right is a visual representation of a sequence of numbers called a Mandelbrot set.

Sequences that follow a pattern can be described in a number of different ways. They may be listed in sequential order; they may be described as a functional definition; or they may be described in an iterative definition.

Listing in sequential order
Consider the sequence of numbers t: {5, 7, 9, …}. The numbers in sequential order are firstly 5 then 7 and 9, with the indication that there are more numbers to follow. The symbol t is the name of the sequence, and the first three terms in the sequence shown are t1=5, t2=7 and t3=9. The fourth term, t4, if the pattern were to continue, would be the number 11. In general, tn is the nth term in the sequence. In this example, the next term is simply the previous term with 2 added to it, with the first term being the number 5.

Another possible sequence is t: {5, 10, 20, 40, …}. In this case it appears that the next term is twice the previous term. The fifth term here, if the pattern continued, would be t5=80. It can be difficult to determine whether or not a pattern exists in some sequences. Can you find the next term in the following sequence?

t: {1, 1, 2, 3, 5, 8, …}

Here the next term is the sum of the previous two terms, hence the next term would be 5+8=13, and so on. This sequence is called the Fibonacci sequence and is named after its discoverer Leonardo Fibonacci, a thirteenth century mathematician.

Here is another sequence; can you find the next term here?

t: {7, 11, 16, 22, 29, …}

In this sequence the difference between successive terms increases by 1 for each pair. The first difference is 4, the next difference is 5 and so on. The sixth term is thus 37, which is 8 more than 29.

Functional definition
A functional definition of a sequence of numbers is expressed in the form:

tn=2n-7, n∈ {1, 2, 3, 4, …}

Using this definition the nth term can be readily calculated. For this example t1=2×1-7=-5, t2=2×2-7=-3, t3=2×3-7=-1 and so on. We can readily calculate the 100th term, t100=2×100-7=193, simply by substituting the value n=100 into the expression for tn.
Look at the following example:

dn=4.9n2, n∈{1, 2, 3, …}

For this example, in which the sequence is given the name d, d1=4.9×12=4.9, and d2=4.9×22=19.6. Listing the sequence would yield d: {4.9, 19.6, 44.1, 78.4, …}. The 10th term would be d10=4.9×102=490.
Here is another example:

cn=cos⁡(nπ)+1, n∈{1, 2, 3, …}

Here the sequence would be c: {0, 2, 0, 2, …}.

Question 1: Find the 20th term of the sequence 2⋅4, 4⋅6, 6⋅8, …, n terms.
Solution:
The given sequence is 2⋅4, 4⋅6, 6⋅8, …, n terms.
nth term an=2n⋅(2n+2)

an=2×20⋅(2×20+2)=40⋅42

Thus, the 20th term of the series is(40⋅42).

Question 2: What is the 20th term of the sequence defined by an=(n-1)(2-n)(3+n)?
Solution:
Putting n=20, we obtain

a20=(20-1)(2-20)(3+20)
=19×(-18)×(23)=-7866.
Q3. Find the indicated terms in each of the following sequences whose nth terms are:
(a) an=5n-4; a12 and a15 (c) an=n(n-1)(n-2); a5 and a8
(d) an=(n-1)(2-n)(3+n); a1, a2, a3
(e) an=(-1)n, n; a3, a5, a8.
Solution:
We have to find the required term of a sequence when nth term of that sequence is given.
(a) an=5n-4; a12 and a15
Given nth term of a sequence an=5n-4
To find 12th term, 15th terms of that sequence, we put n=12, 15 in its nth term.
Then, we get

a12=5⋅12-4=60-4=56
a15=5⋅15-4=75-4=71

∴ The required terms a12=56, a15=71. To find 7th, 8th terms of given sequence, we put n=7,8. The required terms a7=19/33 and a8=22/37.

(c) an=n(n-1)(n-2); a5 and a8.
Given nth term is an=n(n-1)(n-2).
To find 5th, 8th terms of given sequence, put n=5,8 in an then, we get

a5=5⋅(5-1)⋅(5-2)=5⋅4⋅3=60
a8=8⋅(8-1)⋅(8-2)=8⋅7⋅6=336

∴ The required terms are a5=60 and a8=336.

(d) an=(n-1)(2-n)(3+n); a1, a2, a3.
The given nth term is an=(n-1)(2-n)(3+n).
To find a1, a2, a3 of given sequence put n=l, 2, 3 is an

a1=(1-1)(2-1)(3+1)=0⋅1⋅4=0
a2=(2-1)(2-2)(3+2)=1⋅0⋅5=0
a3=(3-1)(2-3)(3+3)=2⋅(-1)⋅6=-12

∴ The required terms a1=0, a2=0, a3=-12.

(e) an=(-1)n n; a3, a5, a8.
The given nth term is, an=(-1)n⋅n.
To find a3, a5, a8 of given sequence put n=3,5,8, in an.

a3=(-1)3⋅3=(-1)⋅3=-3
a5=(-1)5⋅5=(-1)⋅5=-5
a8=(-1)8⋅8=(-1)⋅8=-8

∴ The required terms a3=-3, a5=-5, a8=8.

Q4. Find the indicated term
(a) an=3n; a8
(b) an=3n-5; a16
(c) an=3n;a5
(d) an=(-3)n;a6
(e) an=(-1)n;a50 A series is the sum of the terms of a sequence.
For the finite sequence {un} with n terms, the corresponding series is u1+u2+u3+⋯+un. The sum of this series is Sn=u1+u2+u3+⋯+unand this will always be a finite real number.
For the infinite sequence {un}, the corresponding series is u1+u2+u3+⋯+un+⋯.
The formula how to get nth term for every series is given by

un=SnS(n-1).

In many cases, the sum of an infinite series cannot be calculated. In some cases, however, it does converge to a finite number.

Q5. lf the sum of the first n terms of an arithmetic series is 4nn2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
Solution:
The sum of n terms of an AP is given by Sn=4nn2
Putting n=1, we get
First term =a1=S1=4⋅1-12=3
Putting n=2, we get
Sum of two terms =a1+a2=S2=4⋅2-22=4

a1+a2=4
3+a2=4 [∵ the first term a1=3]
a2=1

Hence, the second term is 1.
Common difference d=a2a1=1-3=-2
Therefore, the tenth term =a10=a+9d=3+9(-2)=-16.
Similarly, the nth term =an=a+(n-1)d=3+(n-1)(-2)=5-2n

Q6. The sum of the first n terms of a series is given by Sn=4nn2.
Find: (a) un(b) u18.
(c) If Sn=-60, find the value of n.
Solution:
(a) Sn=4nn2

S(n-1)=4(n-1)-(n-1)2
=4(n-1)-(n2-2n+1)
=4n-4-n2+2n-1
=-n2+6n-5

un=SnS(n-1)

=(4nn2 )-(-n2+6n-5)
=4nn2+n2-6n+5
un=5-2n
(b) un=5-2n
u18=5-2(18)
=5-36=-31

(c) Given: Sn=-60

∴4nn2=-60
n2-4n-60=0
(n+6)(n-10)=0
n=-6 or n=10
n=10, as n∈N