Arithmetic progression

1
What is arithmetic progression? An arithmetic
progression can be understood as a sequence that
has a starting number and a series advancing
with a definite interval. The simplest and
easiest example being the number series
1,2,3,4,5,6,7. Here, the first number is 1 and
the definite interval is 1. If we keep on adding
1 to the last number, the sequence will keep on
progressing and yes, it becomes an arithmetic
progression! Most used terms in arithmetic
progression lesson Now, let us transform this
definition into a mathematical formula to sound
more studious. Let us name all the terms first.
(Most used terms in arithmetic progression
lesson) When the sign of comes in picture, an
expression becomes an equation. So, the
expression 24x6 becomes an algebraic
equation. First term be a1, Difference be d and
Last term be an Now, add d to a1. What do you
get? (a1d). It means you got two numbers of the
sequence. The two numbers are a1 and (a1d). a1
a1 a2 (a1d) a3 (a1d) d a3 a1
2d Similarly a4 can be written as a4 a1
3d In the same manner, if you need to find any
number of the given arithmetic progression, you
can simply apply this trick, and lo, the
mathematicians call it a formula. Did you guys
realize? You have just made up formula. So, here
goes the formula To find the nth term of any
progression? an a1 (n-1)d
2
Sounds simple right? Well, yes it is. Lets check
it out with the help of one example. My
birthdate is 6 so let the first term be 6. And my
birth month is 3, so let the difference be 3. We
have got a16 and d3. Let us put it into the
formula and find out the 11th term of the
series. an a1 (n-1)d a116(103)
a1136 Bingo! As easy as that. Now that you
have learned how to calculate any number in a
given arithmetic progression, lets proceed onto
the next formula. Sometimes, finding the nth term
is not enough and you need to find the sum of a
definite arithmetic progression. How to go about
with that? The sum of arithmetic progression The
sum of the nth term of any given arithmetic
progression can be calculated by the sum of the
first term and the last term divided by half and
multiplied by the number of terms in the series.
To make it easier, lets put it into a
formula. Snn/2(a1an) But this sounds quite
simple. When we delve deeper into the subject,
you may realize that the nth term is not
necessarily given every time. In that case, you
need to depend upon another formula, that
is Snn/22a1 (n1)d If you are wondering
how we came up with this formula, let me explain
to you. We did nothing but substituted the value
of an from the previous formula. Lets check
out. Snn/2(a1an) But, we know an a1
(n-1)d Let us substitute this value in the Sn
formula.
3
Snn/2(a1an) Snn/2(a1a1 (n-1)d) Snn/22a1
(n-1)d How cool and easy! Let us take a few
examples to understand it better. So, lets
begin with my favorite numbers. Consider the
first sequence where a16 and d3, and suppose
the last value a be 36. Now, putting these in the
first formula of Sn, we will get Snn/2(a1an) S
n11/2(636) Sn231 Bravo! Now, lets proceed
with another example where we dont know what is
the last number of the sequence. Suppose the
value of a13 and d6, we need to find the sum
of the number till the 20th number of the series.
Lets go! Snn/22a1 (n-1)d Sn20/223
(20-1)6v Sn106114 Sn1200 Did you know?
After becoming an expert, you can also calculate
the amount collected in your piggy bank (of
course, if you add money at proper intervals and
dont cheat). Suppose you initially put 500 in
your piggy bank and hereafter keep on adding 100
every month, how much money will you have at the
end of the year? Lets calculate. a1500 d100
n12 an a1 (n-1)d an 500 (12-1)100 an
1600
4
So, you will be getting 1600 by the end of the
year. Good savings, indeed! In a similar
fashion, you can calculate the fare of a taxi
when you travel. They have a fixed initial rate
and the rate increases according to the distance
you travel. So, ask your cab driver the next
time you travel and calculate your own
fare. Amazing! Isnt it? It seems you are
becoming an expert in solving arithmetic
progressions! To find the sum of an arithmetic
progression but from the last term. Lets go
further and cover up the one last formula as
well. It is not any different, just a twisted
version of the same thing. Suppose you need to
find the sum of an arithmetic progression but
from the last term. It means, you need to find
the sum of the arithmetic progression but in
reverse order. Let's try to put it in our
formula for finding the sum. Snn/22a1
(n-1)d But as we have to find it out from the
last term, lets call it al. So, lets replace
a1 with al and as we have to go in reverse order,
the difference d will become (-d). By replacing
all the values, you will get Snn/22a1
(n-1)d Snn/22al (n-1)(-d) As simple as
that. Just in case you need to understand it
better, lets try to find the sum of the first
100 even numbers from the last number. So here,
al 100, d(-2) but we dont know how many even
numbers are there from 1 to 100. So, first, we
need to find the value of n. Lets
start. Formulae known to us an a1 (n-1)d
5
For this problem, we have the following values
with us an100 a12 d2 Putting these in the
equation, well find the value of n. an a1
(n-1)d 1002(n-1)2 982n-2 962n 48n Hence,
there are 48 even numbers from 1 to 100. Now,
lets find the sum from the last
number. al100 n48 d(-2) Snn/22al
(n-1)(-d) Snn/22al (n-1)(-d) Sn48/22100
(48-1)(-2) Sn2,544 See! We have understood
the fundamentals of arithmetic progressions like
a cakewalk. So bring on your witty caps and
solve as many problems you come across while
reading this topic and dont shy away from this
fun-filled, easy, and understandable topic.
Happy solving!