Video: Arithmetic Mean Video Lecture | Mathematics (Maths) Class 11 - Commerce

hi so in this video again we are going
to discuss about inserting n number of
arithmetic means between any two given
quantities and in this video we are
actually going to discuss about the
formulas so the people who like to
remember formulas to find solution of a
mathematical problem this will be very
helpful to them but I usually suggest to
start from the first principles start
from the basics and use your intellect
every time and then everything becomes
very easy I always advise to follow the
basics to find the solution for any
problem but we should know the formulas
also so let's derive the formulas to
insert n number of arithmetic means
between any two given quantities so if
we take any two numbers like a and B and
if we have to insert n number of
arithmetic means between these numbers a
1 a 2 are the arithmetic mean still e n
and B is the last term and if a 1 to a n
are the arithmetic means and if I write
them in between a and B in this manner
then this will form an arithmetic
progression so this is the definition of
inserting n number of arithmetic means
between any two given numbers right the
arithmetic means if we insert it should
form an arithmetic progression right
that is the definition and how can we
insert these numbers to insert these
numbers we have to find the common
difference right we have to find D and
what we are provided with we are
provided with first term as a and we are
provided with last term as B and we are
provided with n also total number of
terms what is total number of terms over
here
if we are inserting n arithmetic means
between a and B then total number of
terms becomes n plus 2 let's call this
total number of terms as N 1 and the
first term as even just to differentiate
then we have a formula for the last term
what is the formula for the last term
l is equal to e plus n minus 1
multiplied by D right and since here we
have represented the first term as even
and the number of terms as n1 I am just
writing in the same notations a1 + n1 it
is the first term and this is the number
of terms right so in order to find the
value of T if I substitute the value of
L as B and even as E and n1 as n plus 2
right then we have this minus 1
multiplied by D right
so from here if I bring this a to the
left of equal to sign it will become B
minus a is equal to n plus 1 multiplied
by B plus 2 minus 1 makes plus 1 right
and from here we will get B is equal to
B minus a divided by n plus 1 right so
if you have to insert n number of
arithmetic means between two terms a and
B then the common difference will be B
minus a divided by n plus 1 so this is
one formula to find the common
difference right so we have got D is
equal to B minus a divided by n plus 1
right and what will be this progression
this progression will be written as
general form of any arithmetic
progression is a a plus D a plus 2 D and
a plus 3 D and this continues right so
if I substitute the value of D as B
minus a divided by n plus 1 this
progression will become a
a plus B minus a by n plus one this will
be the second term and the third term
will be a plus two B minus a divided by
n plus one right and the term will be a
plus 3b minus a divided by n plus one
and this will continue till B right so
this is the generalized form of the
arithmetic progression when you insert n
number of arithmetic means between a and
B so this is the derivation of the
formulas and to keep updated about the
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