Proof of Sum of n terms of AP Video Lecture - Class 10

hi welcome to tutorial where I'm going
to show you how to prove the
SN of a north
progression now what I've got here is
the first three terms in an arithmetic
progression it just in general first
term remember was always a and if the
common difference was D the second term
would have been a plus one D or a plus D
third term a plus 2 D and so on so if
we're looking at the nth term the nth
term would have been a plus n minus 1 D
and the term before that would have been
the N minus 1 term that is if I was to
replace n with n minus 1 then the term
before it would have been a plus n minus
2 D and so on the term before this would
have been a plus n minus 3 D well I'll
leave out those terms and we just put
dot dot and so on now the purpose of
this particular tutorial then is to find
the sum of all these terms which we call
SN so if I write down SN the sum of the
first n terms let's go into equal then
the first term a plus nor leave a space
you'll see why in a moment then we've
got the second term a plus D and then
we've got the third term a plus 2 D and
so on and then we've got the next term a
plus n minus 2 D and then plus the last
term a plus n minus 1 D so write out all
the terms it might be a good idea just
to put brackets around these terms again
you'll hopefully see why in a moment
when we get down here let's just put
square brackets around these ones all
right ok so we can clearly see what the
terms are now what I'm going to do now
is just add the terms up in
a reverse order just write them down
we'll start with this one and work all
the way back down here so obviously it's
it going to be exactly the same value SN
but we'll start with this one so
underneath here we'll write a plus n
minus 1 D and then directly under this
one and that's why I left the space
we're going to write this term that
would be a plus n minus 2 D the next one
would have been this term before this
one here a plus n minus 2 D the term
before it would have been a plus n minus
3 D so write that down on the leg here a
plus n minus 3 D and this would have
gone on and so on plus up to this term
here this would have been underneath
this one a plus D and then we would have
had plus the very first term a let's put
these in brackets just to segregate them
from the next term all right
so hopefully you can see the terms
through here so we'd have had plus and
so on all the way up to a plus D and
then a right now what I'm going to do is
add these two equations together we
should really give them numbers so let's
call this equation number one and this
one equation two so what I'm going to do
is do 1+2 1+2 what does it give us well
SN plus another SN is 2 SN so therefore
we would have 2 SN equals now if I add a
2 a plus n minus 1 D what am I going to
get I'm going to get two A's plus n
minus 1 D so it'd have to a plus and n
minus 1 D now if I add this pair of
terms what are we going to get
well we're certainly going to get two
A's there right and here if I was to
expand this bracket I've got n D minus
2d the minus 2d plus Rd is going to be
minus one D so I've got n minus one D ah
that's the same as what I had here so
I've got another two a plus n minus one
D let's just put that down
plus another two a plus n minus one D
and we'll put that in brackets early
just to say that we've added the first
pair of terms and now we're adding the
second pair of terms and that gave us
that one there let's add this one up
what again we've gotten a plus an A so
that's going to be a 2 a and then we've
got 2 D added 2 n minus 3 D well that's
going to be ND - 3 D if I opened up the
brackets in the - 3 D but then if I add
2 D to that I'm going to have n D minus
1 D again n minus 1 all times D so I've
got another one of these brackets to a +
n - 1 D well it turns out that we're
always going to get this particular term
here when we add corresponding terms
like this you can even see at the end
here look it's going to carry on all the
way down to the very last term because
when you add this pair of terms a plus a
is 2 a and then you just got simply + n
- 1 D n minus 1 or multiply by D and
I'll put that in its own square bracket
there so this calculation this
particular term ends up being written
down is the same number of terms as I
had along here which was in terms so
really what I've got is - lots of SN
equals n times this
do n times 2a plus all of n minus 1 D
now if I divide both sides by this 2
here
what that's going to give me is going to
be SN equals n over 2 all multiplied by
2 a plus n minus 1 times the common
difference D and that was the formula
that I set out to show you don't forget
there's also another one that we can get
as well from this because what I can do
with this formula is write it like this
we'll just do it over here all right
we've seen that SN is equal to n over 2
multiplied by 2 a let me write that 2 a
is a plus another a and then we have the
plus n minus 1 D so plus n minus 1 D now
a plus n minus 1 D that's this bit up
here and this was the last term in the
sequence so I could call this L for
short so what I end up getting is n over
2 multiplied by a that's that a there
and this particular part is say L where
L ok is the last term
so I've been able to show you then the
two proofs of SN the sum of the first n
terms we have this particular formula
for SN and we have this particular
formula for SN so I hope you've been
able to follow the proofs for getting
those two formulae and that brings us to
the end of this tutorial
you