Advanced Algebra : Sequence & Series Video Lecture - Banking Exams

hi students this is Monica Benson from
bullseye I have been guiding students on
various aspects of quantitative aptitude
for last 10 years
today's topic is sequence and series we
are going to discuss some special kind
of sequences that is a section of
advanced algebra however to ensure the
better understanding of the concept it
is imperative to know the basics of
arithmetic progression and geometric
progression so let us have an overview
of the session first of all there will
be questions based on AGP series we call
it arithmetic or geometric progression
then you will have some question based
on some special series then you will
have some practice questions so let us
move to AGP series question says get s
is equal to 2 by 3 plus 5 by 3 square
plus 8 by 3 cube plus 11 by 3 raise to
power 4 plus so enough to infinity we
want to calculate s we want to find the
sum of infinitely many terms as you can
see in the emulator we have an
arithmetic progression what is the first
term here that is 2 what is a common
difference that is 3 and in the
denominator we have a geometric
progression what is the first term of
this GP 1 by 3 and what is a common
ratio that is 1 by 3 again this is our
first term and this is the combination
of the GP so that is a combination of
arithmetic progression and geometric
progression so we call it call then a GP
series are automatically emetic
progression we have a particular method
to solve these kind of sums let us see
the method then we'll come to the
shortcut first of all we will multiply
the whole series the whole sequence by a
common ratio which is 1 by 3 in this
case so it will become
1 by 3s now what will be the first term
in this case it will be 2 by 3 square we
will leave the first place empty we will
write the first term here then take the
subtraction it will become 2 by 3's
first term as it is now 5 minus 2 3
since we have same denominator now you
can see that the second term onwards we
have a pure GP we are drawn with clique
progression and there are infinitely
many terms in that GP we have a formula
to calculate sum of infinitely many
terms of a GP what is the formula that
is a upon 1 minus R where R stands for
first term and R stands for common ratio
so we can write 2 by 3 s is equal to
first term again as it is 2 by 3 plus
now we will apply the formula so it will
become 3 by 3 square that is a upon 1
minus R which is 1 by 3 in this case if
you solve it you will get s is equal to
7 by 4 I am only doing the calculation
part for you so this is the answer in
this case 7 by 4 now come to the
shortcut this is direct formula for
infinite sum of an AGP it says as
infinity and in sum of infinitely many
terms this is equal to a B upon 1 minus
R plus DB R upon 1 minus R whole Square
where a is the first term of the a B B
is the first term of the G B D is the
common difference of the a P and R is
the common ratio of the G V let us apply
the shortcut I'm taking the previous sum
only so if we apply this formula as
infinity will become a into B divided by
1 minus R Plus D be R upon 1 minus R
whole square so if you solve it you will
get the same answer which turns out to
be 7 by 4 now let us move further there
is one more question based on AGP CDs
again we want to find the sum 1 by 5
plus 3 by 5 square plus 6 by 5 cube plus
10 upon 5 raised to power 4 plus 15 upon
4 5 raised to power 5 and so on up to
infinity in the numerator if we take the
difference here the difference is 2 then
we have a difference of 3 not the same
difference
for 5 so that is not a direct version of
AGP so we cannot apply the direct
formula for refining the infinitely many
terms some of AGP but in the second step
if we see it becomes an arithmetic
progression so how can we solve these
kind of sums we'll apply the same
technique let s is equal to 1 by 5 plus
3 by 5 square plus 6 by 5 cube plus 10
by 5 raised to power 4 and so on up to
infinity what is the common ratio here
the common ratio is 1 by 5 so I am going
to multiply the whole series with a
common ratio so the first term will
become 1 by 5 square again leave the
first place we will write 1 by 5 square
here 3 by 5 Q 6 by 5 raised to power 4
and so on up to infinity
then we will take the subtraction
subtraction will become 4 by 5 s first
term as it is now what does since the
denominators are saying we can take
subtraction of numerators only it comes
out to be 2 by 5 square then it will
become 3 by 5 Q then it will become 4 by
5 raised to power 4 sewn up to infinity
again you can see in the numerator we
have our direct AP so we can apply a
formula here to apply the formula we
will write the first term of AP which is
1 what is a common difference that is 1
again what is B B means first term of
the GP it is 1 by 5 and what is a common
issue that is 1 by 5 again so we will
apply the formula here which comes out
to be a B
upon 1 minus R + DB R upon 1 minus R its
whole square so if you solve it you will
get the answer again I'm leaving the
calculation part S comes out to be 25 by
64 and that is the final answer in this
case now let us move to next question
again a new kind of series we want to
find the sum of the series 1 upon 2 into
5 plus 1 point 5 into 8 plus 1 upon 8
into 11 Plus soon up 2 plus 1 upon 35
into 38 there is a pattern going 1 if
you observe what is the difference every
time this is 3 1 1 2 into 5 2 plus 3
will become 5 5 plus 3 will become 8 and
so on so how to solve these kind of
firms again there is a method in this
case you will take the difference as you
can see this is 3 so we will multiply
and divide the whole series by 3 I mean
it will become 1 by 3 we are dividing it
by 3 and we are multiplying it by 3 so
it will become 3 on 2 into 5 plus 3 upon
5 into 8 plus so not 2 3 upon 35 into 38
what is the usage of this see now this 3
can be written as 5 minus 2 it can be
written as 8 minus 5 as you can see here
we have 5 into 8 that's why I have
written this as 8 minus 5 since we have
in denominator 38 into 35 that's why we
are writing it as 38 minus 35 so if we
take the very first term
it will become 5 minus 2 upon 2 into 5
if we separate them it will become this
5 upon 2 into 5 - 2 upon 2 into 5
similarly the next term will be 8 upon 5
into 8 minus 5 upon 5 into 8 the last
term will become 38 upon 35 into 38
minus 35 upon 35 into 30 now see the
magic what is the usage of all these
things now it has become 1 upon 2 minus
1 upon 5
they got cancelled plus 1 upon 5 minus 1
upon 8 plus 1 on 8 minus 1 upon 11 the
last term will become 1 upon 35 minus 1
upon 38 now if we take the second term
of first bracket and the first term of
second bracket they got cancelled so all
the term will get cancelled except the
first and the last term and this will
happen always so my answer will become 1
by 3 1/2 minus 1 upon 38 if we solve it
we will get 3 by 19 that will be the
final answer in this case so directly we
can learn we can not assume that
whenever the last term is mentioned we
will take only the first term and the
last term so my answer will become 1/2
minus 1 upon 38 into 1 upon the
difference what is the difference here 3
will be the difference here so that will
be the final answer in this case 3 by 19
this is the answer
now let us move further a huge question
it says there are eight four three six
balls each with a radius of one unit
stacked in a pile with one ball on top
three balls in second air six in third
layer 10th in fourth year and so on find
the number of layers in the pipe so let
us understand the meaning of this
question we have some balls in a pile we
have one in the first file then three in
the second layer then six in the third
layer and 10 in the fourth layer and we
don't know the number of layers here
that is a question so let us say there
are n layers so I'm writing this as n
terms and this total comes out to be
eight four three six we want to
calculate n here so let us see we will
take the help of series again it says 1
3 6 10 so not to n terms this is not a
direct ap GP or a GP series but we can
observe that the difference here is 2 3
4 and shown up how can we solve these
kind of series we will try to write the
nth term of these kind of series as we
can see that T 1 is 1 here I'm in the
first term T 2 is 3 here that is 1 plus
2 we are adding 2 to first of then we
have T 3
that is one plus two plus three then we
have T 4 which is 1 plus 2 Plus 3 plus 4
so by gentle izing we can say that T and
will be nothing but summation and see
the first term we are taking only 1 the
2 of the second term we are taking 1
plus 2 and in the third term we are
taking sum of all natural numbers up to
3 in the fourth term we are taking some
of natural numbers up to 4 so we can
generalize that in T and we will take
some of in natural numbers so it has
become Sigma and we have a formula for
this this is heading to n plus 1 by 2
that is our system now we want to write
a severe sum of n terms so how can we do
this and it term can be written as half
into n square plus n so if we apply
Sigma here it will become half Sigma n
square plus Sigma n how come let us see
CRT n is half N squared plus n what will
be T 1 if we write in this manner it
will be 1 square plus 1 what will be T 2
it will be half of 2 square plus 2 and
so on up to TN it will be half of N
squared plus n we want to calculate SN
what do you mean by SN that is sum of
first n terms so if we combine them we
get 1 square plus 2 square plus so on up
to n square that is the first column
plus now the remaining things 1 plus 2
plus one up to n
so it comes out to be half of Sigma n
square plus Sigma n let us solve this
what is Sigma n square formula it is n
into n plus 1 into 2 n plus 1 divided by
6 plus what is Sigma n formula n into n
plus 1 divided by 2 so first of all the
base take n into n plus 1 divided by 2
common we'll be left with 2 n plus 1
divided by 3 plus 1 which will become n
into n plus 1 divided by 4 into 2 n plus
4 divided by 3 if we solve it the final
expression will become n into n plus 1
into n plus 2 because 2 will be common
here divided by 6 now this sum of n
terms if we are considering that there
will be n layers this is given as 8 4 3
6 so if we solve it we get the answer we
can go by options by taking options we
can see that the second option is the
answer so there will be 36 layers in
that file let's move further a nice
question on Fibonacci series it says
find the number of all 10 digit natural
numbers made by using digits 3 and 5
only such that no two threes are
consecutive no two T's are together so
what is our conventional approach we'll
try to make cases since it says that no
two threes are together consecutive so
we can take number of files and and
there will be no three how many such
numbers are possible there is only one
number that is all files then we will
take the second case the number of files
are 9 and there is only 1 3 so 3 can
occupy
any of the 10 places that's why there
are 10 numbers possible if we keep on
going in such manner we will have many
cases and by adding all of them we will
get 144 as the answer but it is
cumbersome so we will make use of a
beautiful series which is known as
Fibonacci series see we'll go by taking
smaller length of the number let us take
there is only one digit number so I am
applying the same condition that we want
to make numbers by using only three and
five and no two threes are transitive
so we are talking about one digit number
so how many numbers are possible only to
either it will be three or it will be
five so in total how many numbers are
possible to what about if I take two
digit numbers so if we make cases since
our condition is no two threes are
consecutive number can be 55 it can be
53 it can be 35 but it cannot be 33
because of the condition given so how
many such numbers are possible they are
three in total so we have seen if we are
talking about one digit of such number
they are two in number if we are talking
about two digit numbers they are three
in total what about if we take three
digit numbers either my number will
start with a five or my number will
start with a three and there are two
places remaining so if we take the first
case in which the number is starting
with five remaining two places on
nothing but all the 2 digit numbers
which are satisfying the same condition
and we have calculated that that there
are three such numbers once again if we
are starting on
by five either my number will be 55 555
or it will be 553 or it will be 535 so
these are nothing but the caves which we
have calculated previously now let us
come to the second case where my number
is starting with a three after three you
cannot take a three because of that
condition that no two threes are
consecutive so we have to take a five on
the second case so the remaining numbers
are nothing but all the one digital
number which are satisfying this
condition and we have calculated it
previously so how many such 3 digit
numbers are possible they are 5 into 10
once again if we talk about four digit
numbers either my number will start with
a 5 or my number will start with a 3
since we are talking about 4 digit
numbers three places are remarried so if
we take the very first case where the
number is starting with a 5 we still
have 3 places to fill so these 3 places
are nothing but all the 3 digit numbers
which are satisfying the same conditions
and we have calculated those in the
previous case they are 5 in total
similarly if we take the next case where
our number is starting with the three
after 3 you cannot take a 3 that's why
we will put a 5 here so we are talking
about all the 2 digit numbers which are
satisfying the same condition that we
have calculated previously so it would
be nothing but in total 8 so every time
we are calculating we are increasing the
length of the number suppose we are
calculating 5 digit numbers it will be
nothing but some of previous two places
so we can say that if we talk about one
digit in number they were 2 we talk
about 2 digit number they were 3 3 digit
that is nothing but the total of
previous to turn that is five forward
edited again the sum of previous two
terms that is eight three plus five if
we go on in this manner we can calculate
ten digit number which comes out to be
144 and that was the answer in this
question so this kind of series is known
as Fibonacci series
very very interesting and important
series where the N it term is sum of
previous two terms only first two terms
are defined in the series and the third
term onwards we can write the whole
series mix every done is nothing but sum
of previous winters so applying this
technique we can calculate the number of
all ten digit numbers in a jiffy because
we have to calculate a series only not
so many cases so answer is 144 in this
case now these are some practice
questions
and these are the answers for more
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