Complex numbers

“Mathematics is the queen of science and arithmetic is the queen of mathematics.”  These were the words of Carl Gauss, the person who gave the world sequence and series arithmetic progression concept along with several other major contributions in the field of mathematics. Everything around us follows a fixed pattern be it the structure of leaves or the railway tracks and so does everything around us. We have been observing and following the patterns, sequences since our childhood and Carl Gauss was able to understand and apply this knowledge of sequence and solved a lengthy math’s problem in his school kids which is today known as arithmetic progression. Arithmetic progression is nothing but a sequence in which a constant difference (fixed number) which can be either positive or negative is added to any number except the first number to obtain the next number. 

Sequence and series

• Consider a series below

1, 2, 4, 7, 11, 16, 22 …. ——- (1)

5, 15, 35, 65, 105, 155…. ——–(2)

1, 0, 1, 0, 1, 0… ——-(3)

In eq (1) we can see that the series is progressing with +1, +2, +3, +4 and so on 

Likewise, in eq (2) we can see that it is progressing with +10, +20, +30 and so on 

Similarly, in eq (3) we see that it is following a pattern of -1, +1  

• Now readers consider another series below

1, 2, 3, 4, 5……… ——(4)

1, 3, 5, 7, 9………… —-(5)

5, 10, 15, 20, 25………. —–(6)

In eq (4) the difference between the terms is 1 i.e. you add 1 to each number to get the next consecutive number

Similarly, in eq (5) if you add 2 to the previous number you get the next number

And a similar case follows with eq (6)

Now if we observe carefully, we can say that all the above equations do follow some pattern and we can happily define a sequence from this understanding that a sequence is the collection of numbers arranged in the definite order according to some definite rule.

Also, when we look at eq (4 to 6) we find that the difference between two consecutive terms is the same , which is called progression. Every progression is a special sequence.

So now coming back to our definition of arithmetic progression (A.P), A.P is a special type of sequence where the difference between the two consecutive terms is always the same/Constant.

And this difference is said to be common difference and is denoted by d

General representation of A.P

Consider this below sequence which is in A.P

1, 2, 3, 4, 5 ………n

Here the first term is 1, which we will denote as t1

t2 = second term = 2 

t5 = fifth term = 5

tn = nth term

to find the common difference we do t2 – t1 or we can say we would do tn – tn-1

i.e., 2 -1 =1 is the common difference(d) here

Let’s try to formulate the formula for A.P using our current understanding of our topic.

If t1 is denoted as a ( first term) and common difference as d then 

a+d = t2 

t3 = t2 + d = a + d +d = a +2d 

t4 = t3 + d = a +3d.

tn = tn-1 + d = a + (n-1)d

and the sum of AP is given by SN = N2[ 2a+N-1d]

 Application of the above formula

1. Find the terms which are in A.P

• 2, 4, 6, 8…..

This is an example of infinite sequence

Here t1  = 2 , t2 = 4 , t3 = 6 ,  t4 = 8

t2 – t1 = 4 -2 = 2 

t3 – t2 = 6 -4 = 2

Since the common difference is constant the given sequence is an A.P.

• 2, 5/2, 3, 7/2

t1 = 2, t2 = 5/2, t3 = 3, t4 = 7/2

t2 – t1 = 5/2 -2 = ½

t3 – t2 = 3 – 5/2 = ½

This is also A.P.

1. Write the A.P if a = 10, d = 5

Solution: t1 = 10 

t2 = 10+5 = 15

t3 = 15+5 =20

t4 = 20+5= 25

so, the A.P is 

10, 15, 20, 25……

1. 7, 13, 19, 25 … find the 19th term

Solutions: t1 = 7, t2 = 13, t3 = 19, t4 =25

Common difference (d) = 13 – 7 = 6

We know that, tn = tn-1 + d = a + (n-1) d

t20 = 7 + (19-1) *6 = 7 + 18*6 = 7 + 108 = 115

1. Find the number of 3-digit numbers divisible by 5

Solutions: the smallest 3-digit number is 100 and the largest is 999

The numbers which are divisible by 5 in this range are 

100 ,105, 110 ……..995 

So, first term = a = 100 and last term = 995 and d = 5

Using our A.P formula we get, 

995 = 100 + (n-1) *5 

995 -100 = 5 (n-1) 

895/5 = n-1 

179 = n-1 

N = 180.

1. If t11 = 16, t21 = 29, t41 =?

Solutions: using the A.P formula we can write 

16 = a + (n-1) d where all variables carry their usual meaning

16 = a + (11 -1) d = a + 10d —–(1)

29 = a + 20d ——(2) 

Solving both equations, we get, 

16 -10d = 29 -20d 

-13 = -10d 

d = 13/10; a = 3 

therefore, t41 = 3 + 40 * (13/10)

= 3 +13*4 = 3 + 52 = 55

1. In the natural numbers from 10 to 250, how many are divisible by 4 ?

Solutions: the first number divisible by 4 in that range will be 12. 

250/4 gives remainder 2. That means the last number divisible by 4 is 248. 

Using the A.P formula we get, 

248 = 12 + (n-1) 4 

236/4 = n -1 

59 = n -1 

N = 60

Conclusion

A.P is a special type of sequence where the difference between the two consecutive terms is always the same/Constant. Readers must acknowledge the beauty of A.P and understand that this series can upto infinity.