Progression - Basics Video Lecture - Reasoning for LIC AAO (English) - Banking Exams

hello friends today we are going to
learn an important topic from algebra
and desk topicals progressions now
progressions word is made from the world
progress which means slowly and slowly
increasing the value or it may be
decreasing the value to get some new
value now before moving further let us
take an overview of the topic
progressions now students in this topic
we are going to cover the different
types of progressions now these
different types of progressions are
arithmetic progression and geometric
progression so friends we are going to
look how to identify or recognize
whether a progression is arithmetic
progression or it is a geometric
progression then further we are going to
look at different ways or the formulas
to calculate the energy term as well as
the sum of n terms all the infinite
terms for an arithmetic progression or
geometric progression and then
afterwards we have some basic questions
for you to practice at the convenience
of your home so let us start with the
first part arithmetic progression no
friends how to identify whether a
progression is arithmetic progression or
not so for that
let us take an example of arithmetic
progression
so I'm taking an example in which I am
writing the terms of an arithmetic
progression these terms are 2 5 8 11 14
17 and so on so can we look that the
difference between the consecutive terms
in each case is 3 now this difference
which is constant for the consecutive
tones is called common difference now
this common difference is represented by
a symbol small D which in this case
becomes 3 now students this is very very
important that we always have to find
out the difference of the term with the
preceding term so I can say the
difference will be 8 minus 5 or this can
also be written as 11 minus 8 and so on
further we are representing the first
tone with a so if towards the first term
a then can we observe the second term
file can be made by adding the common
difference with the first term a so 5
can be written as 2 plus 3 similarly can
we look it can be written as 5 plus 3 or
which can be further become 2 plus a
plus 2 D now friends can we say the
third term will be similarly a plus 3 D
and so on this will help us to write the
inner term of an arithmetic progression
so now can we observe a basic pattern
that whenever we are writing a town for
an arithmetic progression this always
includes some coefficient of the common
difference so can we observe when we are
writing the second term the coefficient
of D is 1 when we are writing the third
term the coefficient is 2 and so on so
friends we can write the in it down of
an
netic progression as the first term plus
n minus 1 into the common difference D I
hope this formula is clear ok
now from here we can check that in a
storm of an arithmetic progression can
be written as a plus n minus 1 into D
where P n becomes the inner term is
called the first term and D is the
common difference between the
consecutive terms now friends this is
very very important to learn over here
that in case we want to find out the
last term of an arithmetic progression
we can again use the same formula as we
are going to calculate the inner term so
which means the formula to calculate the
last term or the inner term is the same
now let us take an example to clear this
formula so in this question we have to
calculate the 25th term for a given
arithmetic progression now friends can
we observe that the first term in this
case is 12 while the common difference
can be calculated as any Tom - the
preceding term so which comes out to be
six in the skips now let us write the
formula to calculate the N it's down
this is a plus n minus 1 into D now
putting in the values for the various
variables we can write the twenty-fifth
term is 12 plus 25 minus 1 into the
common difference 6 which upon solving
can be simplified as so this becomes 24
into 6 144 and the 25th term comes out
to be 156
I hope this question makes clear the use
of the inner term formula now friends
let us move on to the next formula to
calculate
some of in terms of an arithmetic
progression now friends let us learn a
new formula for calculation of the sum
of in terms of an arithmetic progression
this formula is sum of n terms can be
written as the number of terms divided
by 2 into 2 a plus n minus 1 into the
common difference D now let us try to
apply this formula for the calculation
of sum of n terms of an arithmetic
progression so students we have to
calculate the sum of 30 terms for this
given arithmetic progression now as
calculated earlier we can write the
first term is 12 while the common
difference for the progression can be
calculated as 6 so now let us write the
formula for sum of n towns as n by 2
multiplied by 2 a plus n minus 1 into D
now putting in the values of various
variables so sum of 30 terms can be
written as 30 divided by 2 multiplied by
2 into 12 plus number of terms 30 minus
1 can be written as 29 and the common
difference 6 further solving it we can
write 15 into 24 plus 29 into 6 is 174
which on simplification gives 15 into
198 and on further simplification gives
us the sum of 30 terms of the
progression as 2 9 7 0 no friends let us
learn another formula for the sum of n
terms of an arithmetic progression this
formula can only be applied if we know
or we can calculate the last term of an
arithmetic
aggression this formula says sum of n
terms is number of terms divided by two
and further multiplied by the sum of the
first term and the last term now let us
take an example to understand this
formula so in this question we need to
find out the sum of all the multiples of
seven which are lying between 43 and 349
so forty-nine will be the first multiple
of seven greater than forty three
similarly 56 63 and seventy are another
further multiples of seven now to find
the last multiple of seven in this
progression can we observe that 350 will
be a multiple of seven because 35 is a
multiple of seven now to get a number
which is a multiple of seven as well as
less than 349 we subtract seven from 350
this will give us 343 which will be
surely a multiple of seven now let us
find out first of all the number of
tones for this arithmetic progression
the sum the formula for the energy tone
of a progression can be written as a
plus n minus 1 into D which is also
applicable to get the last term but in
this case we know the last term is 343
while the first term is 49 we have to
calculate the number of terms so this is
n minus 1 multiplied by the common
difference 7 so after solving we can get
the number of tones to be 43 now further
we need to find out the sum of these 43
terms so can we write this formula sum
of 43 terms will be the number of terms
divided by 2 into the first term
the last comment which upon solving
becomes 43 upon 2 into 392 on further
simplification which becomes 43 into 196
and further it gives us the sum of 43
terms which are multiples of 7 as 8 food
to wait I hope students this question is
here to all of you know students let us
move further to the different type of
progressions called as geometric
progression let us take an example to
understand the basic geometric
progression so we can have 2 6 1854 162
and so on so students can we observe
that 2 multiplied by 3 becomes 6
similarly 6 multiplied by 3 becomes 18
and 18 x 3 becomes 54 and so on now in
this case this multiplication factor is
constant which is called as the common
ratio this common ratio is represented
by the symbol small R which in this case
comes out to be 3 now again similar to
arithmetic progressions we calculate the
common ratio as the ratio of the town
with the previous dot now further if we
write the first term to be a can we
write the second term as 2 multiplied by
the common ratio 3 or in notation that
becomes in Tawaf similarly the third
term can be written as 6 multiplied by
the common ratio 3 so now this can be
written as a R square and similarly the
fourth term becomes a RQ
fifth thumb becomes a our food and so on
now gain students can we observe the
second term has our power one while for
the third term there is power to four
are similarly for third term and fourth
term we have the power one less than the
number of tongue so now we can write the
next term of a geometric progression as
a multiplied by R raised to the power n
minus one now let us take this formula
once again with the help of this
information so we can write for a
geometric progression the energy term
can be calculated as a into R raise to
power n minus 1 where T n represents the
energy term a represents the first term
and R represents the common ratio for
our geometric progression now let us
take an example to learn this formula
for this question we need to find out
the eighth term of the given geometric
progression so now can we write the
first term in this case is 5 while the
common ratio can be calculated as the
second term divided by the first term
which comes out to be 2 in this case now
the formula for the energy term is a R
raised to power n minus 1 so which upon
putting the values becomes a stone is 5
multiplied by the common ratio 2 raised
to the power 8 minus 1 or which upon
further simplification becomes 5 into 2
raised to power 7 all this becomes 5
into 128 which comes out to be 640 so
friends the eighth term in this
geometric progression comes out to be
640 let us move on to
further formula for calculation of the
in terms of a geometric progression now
we have two formulas in this case to
calculate the sum of geometric
progression depending upon whether the
common ratio is greater than 1 or it is
less than 1 so the sum of n terms of a
geometric progression if common ratio is
greater than 1 it becomes a multiplied
by R raised to power n minus 1 upon R
minus 1 whereas for the common ratio to
be less than 1 the formula converts to a
into 1 minus R raised to power n upon 1
minus R again let us take an example to
learn these formulae as well so we have
to calculate the sum of n terms for this
given geometric progression now again
can we see that the first term is 5
while the common ratio can be calculated
as 2 now can we observe this common
ratio is greater than 1 so now writing
the formula for sum of n terms as a by 2
a into R raised to power n minus 1 upon
R minus 1 now putting in the values of
the various variables sum of 10 terms
can be written as the first term
multiplied by 2 raised to power the
number of terms 10 minus 1 upon 2 minus
1 which upon further solving becomes 5
into 1 0 to 3 divided by 1 which gives
us the value of sum of 10 terms as 6 1 1
5
five one one five no friends we also
have the formula for the calculation of
sum of infinite terms of a geometric
progression now this formula is
applicable only when the common ratio is
less than 1 this formula is a upon 1
minus R the common ratio again let us
take an example to learn this formula we
have to find the sum of infinite terms
of the series 60 30 15 7.5 and so on so
in this case the first term is 16 while
the common ratio can be calculated as
the second term divided by the first
term which comes out to be 1 by 2 so can
we observe friends that this common
ratio is less than 1 which means we can
apply the formula for the sum of
infinite terms which was sum of infinite
terms --is a upon 1 minus R the common
ratio so by putting in the values we can
write the sum of infinite terms will be
60 upon 1 minus 1 by 2 can be further
written as 60 upon 1 minus 2 which upon
solving gives us the sum of infinite
terms as 120 I hope students this makes
clear the basic concepts of the various
types of progressions arithmetic
progression as well as the geometric
progression now further you have some
practice questions to solve at the
convenience of your home
and this is the answer T for the
practice questions thank you for
listening to this video