03 - Arithmetic Progressions_Problem Solving_1 - Class 10 - Maths Video Lecture

hello everyone in the previous session
we have learned the basic concept of
arithmetic progression where we learn
that the sequence of numbers follow
arithmetic progression only if the
difference between consecutive numbers
is constant which we call as common
difference and this common difference
may be positive or negative that means
there will always be a number at the
start and then we keep on adding common
difference we get the next number of the
sequence so we can say that the first
term and the common difference are the
main pillars of any kind of arithmetic
progression and here we must note that
if the sequence is increasing the common
difference is positive and if the
sequence is decreasing common difference
is negative and previously we have also
learnt two basic tools one is for
obtaining any term of the sequence and
other one is for calculating summation
of all the terms of the sequence and in
this session we will focus on first
Formula only if we notice here there are
total four parameters D 1 and D and T n
for understanding this let us consider a
sequence the sequence is 2 4 6 8 and so
on so here we can notice the sequence is
starting with 2 so this is our first
term so T 1 is 2 and the difference
between any consecutive number for
example difference between 2 & 4 4 + 6 +
6 + 8 is 2 so the common difference D is
2 or we can even say that the value of
the term at the first position is 2 at
the second position is 4 and like that
only the value of term at the fourth
position is 8 that means we can say that
and represents the position of the term
we're T and represents the value of Tom
at that particular position okay so now
we know what exactly these parameters
represent or denote but many a times we
find it difficult to identify what is
given and what exactly we need to find
so let us learn how to go about it
here is a situation it says Stephie's
salary in the first month is 15,000 and
every month she gets an increment of 100
so what do you understand by it see here
if the increment is of 100 so in the
next month she will get fifteen thousand
one hundred and then fifteen thousand
two hundred and so on so we can say that
the fifteen thousand is the first term
that is t1 and the increment is of 100
that means common difference D is
hundred and further it says we need to
calculate her salary in 20th month so
now tell me what is given and what is
asked the position of the term is given
20 that means N equals 20 and we have to
find the value of the term at that
particular position that is T 20 so we
have got three parameters T 1 D and n
now we can easily find out the fourth
parameter that is T n and it comes out
to be 16900 so her salary in 20th month
will be sixteen thousand nine hundred so
if we understand the situation we can
solve any kind of rithmetic probation
very easily now let me twist the
question a little bit
so you tell me in which month her salary
will be eighteen thousand six hundred so
can you guess what is given and what is
asked C P one and D will remain same but
this time value of the term is given
that is TN equals eighteen thousand six
hundred and the position of that term is
asked so we have to find n alright so by
substituting values we will get n equals
37 that means she will get eighteen
thousand six hundred in 37th month so
now we have understood what is n + TN in
any given situations so let us move to
next situation here it says third term
of an arithmetic progression is 22 that
is T 3 equals 22 and 17 the term is
minus 20 and we have to find seventh
term that means T 7 we know that any
term of the sequence can be written in
the form of first term and the common
difference 77 can be written as t1 plus
6d that means if we have to find the
value of T 7 we should know t1 and D and
we have to find t1 and D by given two
values that is T 3 and T 17 so what do
we do we will write T 3 and T 17 in the
form of T 1 and B only so we will get
two equations in two variables and we
all know how to solve these simultaneous
equations so after solving it we will
get value of D equals -3 and if we put D
in any of the equation we will get value
of T 1 that is equal to 28 that means we
have got our two main pillars T 1 and D
now by substituting the value of T 1 and
D we will get T 7 equals 10 so friends
what do we learn by the solution of this
question we learn that if any two terms
of an arithmetic progressions are given
then not only we can find first-term t1
and the common difference D but also we
can find any term of that particular
arithmetic progression ok so now suppose
if I ask you to find TN instead of t7
what will you do see previously it was
t7 that means instead of n we had to put
7 so wherever there was n we put 7 over
there but now it is asking us to find TN
only so t1 and D will remain same and
we'll write n as it is so by putting all
the values in the formula we will get TN
equals 31 minus 3 M which is a general
expression in the form of n for example
if you want to cross check whether t7 we
found was correct or not so you just put
the value of N equals 7 and you will get
t7 equals 10 similarly you can put
different values for n and you will get
different terms of the sequence so
friends by now you can easily identify
the different parameters of any kind of
arithmetic progression and you can solve
any question asked on that ok so next
time we will try to solve some questions
based on second formula till we meet
next time here is a question for you
find the fifteenth term of the
arithmetic progression from the last
term here is a very small hint for you
that you write the rhythmatic
progression in reversed order and then
solve it