01 - Introduction to Arithmetic Progressions - Class 10 - Maths Video Lecture

hi friends in this session we are going
to discuss about some interesting
patterns friends there are many er
things around us which follows some
specific patterns in the example of this
stair the height of the first step from
the left-hand side is 3.5 feet the
second step is 3 feet third step is 2.5
feet and the last step is to it means
every next step is getting reduced by
0.5 feet here is 4 more example and then
deposited thousand rupees in a bank and
it gives hundred rupee as an interest at
the end of the year so at the end of the
first year the accumulated amount will
be eleven hundred at the end of the
second year it will be twelve hundred
then thirteen hundred and so on so every
time one hundred rupee is added to the
previous amount in this session we are
going to discuss about the patterns
where the succeeding number is formed by
adding a constant number to the previous
number let me share with you a fresh
example of such pattern I have got a new
project by a company and I have to
complete this project within 30 days you
can consider this table as a table of 30
days and the pattern in this example is
the mode of payment that is on first day
of the project I would get $10 and on
second day it will be increased and I
would get $50 on third day $20 on four
today $25 and so on here the mode of
payment is in such a way then the
succeeding payment is incremented by $5
you can consider this table as month of
November here every number is increased
by $5 that is it is progressing by $5
and since we are dealing with the
numbers therefore such pattern is called
as arithmetic progression the short form
of arithmetic progression is AP on 21st
November that is on 21st day it's my
friend's birthday and I have to buy a
gift for him but exactly I want to find
out how much payment I would get on 21st
day of this project so accordingly I can
buy some gift for him but going this way
that is adding $5 to the previous
payment and get their next payment seems
quite lengthy I want to find it in a
very quick and easy way and I know that
every missing things becomes easy with
the help of observations so even I am
going to observe this and find some easy
way to know the payment on 21st remember
that is on 21st day let us see on day
one I am getting $10 I am NOT putting
dollar sign over here then on day two I
am getting increment of five that is 15
on day three
now the increment is doubled there is 5
into 2 and it would become 20 similarly
on day 4 the increment is 3 times and I
am getting 25 now we can observe here on
day 2 the increment was 5 so we can
write 5 as 5 into 1 and similarly on day
1 we can write it as 5 into 0 that will
give me 10 only so observe here on day 1
increment was 0 on day 2
there was one increment on day 3
increment walls doubled and on for
increment a three times so we can say
that on day 21 that is on 21st remember
the increment will be 20 times that is
10 plus 5 into 20 and it will give me
110
friends let's generalize it so that we
can find out the payment on any day let
us denote the payment on first day as
first term that is t1 where T is term
and suffix 1 is the first term and the
second payment as t2 third as t3 then T
4 T 5 and so on up to T 30 so we can
write t1 as 10 t2 as t1 plus 5
where t1 is the first term that is 10 t3
s t1 plus 5 into 2 T 4 T 1 plus 5 into 3
so if we are generalizing it for any
number for any term let us denote it by
p.m. where T is the term where n is any
number we can write it as T 1 plus 5
into n minus 1 where 5 is the common
difference let us denote common
difference as D now we can write general
expression as TM equals t1 plus n minus
1 into D where t1 is the first term and
is the number of terms and D is the
common difference between any two
numbers let's use this equation and try
to find out the payment on 30th day that
is on 30th November so by using this
formula we will write T 30 instead of M
will write 30 and T 1 plus now your n is
30 minus 1 into the common difference T
so T 30 will be equal to T 1 that is 10
plus 29 into the common difference is 5
so the payment on 30th November
we'll be 155 so on 30th day that is on
30 November I will get 150 $5 here we
have got the general expression for any
kind of arithmetic progression by using
this expression we can find out any term
of that progression now let's go to the
next step and try to find out the total
revenue I would get by this project that
is the summation of all 30 payments now
we want to find out the addition of all
the 30 payments that is 10 I am NOT
writing the sign of dollar here 10 plus
15 plus 20 plus 25 plus and so on to 155
so friends can you tell me the total
amount in offense a similar kind of
question was asked to a boy in a school
that is what is the addition of first 10
counting numbers that boy answered this
question very quickly let us see how
what did you do he write this expression
as s is equal to summation of all 10
counting numbers and then again he wrote
this expression but in reversing order
then he added these two expressions now
s + S will give you 2 s 10 plus 1 is 11
9 plus 2 is again 11 and every time you
will get 11 so we will get 11 10 times
since there are 10 numbers we can write
s as 11 into 10 divided by 2 that is by
getting 2 on the right hand side now two
5s a 10 and 11 into 5 is 55 so he got
this value as 55 we are going to apply
the same logic to find the addition of
all 30 payments let's do it
let us write as 30 where s denote this
summation and
30 denotes the summation of 30 numbers
now we can write it as as 30 in
reversing order that is 155 plus 150
plus and so on 20 plus 15 plus 10 let's
add them we'll get 2 times s 30 and the
addition of 155 plus 10 is 165 again 150
plus 50 will give us 165 so we will get
165 how many times so we can write 2
into s 30 equals 165 into 30 and next
step will give us 165 into 30 divided by
2 so by cancelling out we will get 2 4 7
5 it means the total revenue by this
project is 2,400 and $75 now there will
be a question in your mind that can be
generalize it also and answer is yes
let's do it
let's try to generalize it we can also
write the last expression as s 30 equals
number writing 165 as 10 plus 155 you
can also write it as 150 5 plus 10 into
30 divided by 2
when 10 is the first term and 155 is the
last term that is T 30 divided by 2 into
30 so we will write as n is equal to
first down that is T 1 plus DN that is
the last term divided by 2 into total
number of terms by using this expression
we can find out the summation of any
arithmetic series we will not stop here
because we can generalize it
further we know that TM is equal to t1
plus and minus 1 into D so instead of TM
in this expression we can write t1 plus
n minus 1 into D and we will get s n is
equal to t1 plus again t1 plus n minus 1
into D let's take 2 out of the box into
n by 2 so the final expression will
become T 1 plus T 1 that is 2 times T 1
plus n is the number of terms minus 1
into D that is the common difference
into the total number of terms divided
by 2 so if we have to find out the nth
term of any arithmetic progression then
we will use this formula and if you have
to find out the summation of all the
terms then we can use one of these two
formula both will give you the same
answer if you are using this formula
first of all you will have to find out
the value of T n then you can go further
where in this case you can directly use
the formula if you know the first term
that is t1 number of terms and the
common difference that is T so use this
formula to find the anti term of an
arithmetic progression and you can use
any of these two formula to calculate
the summation of n terms let's have a
recap of this session
you
I hope you liked this video if you have
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