Arithmetic Progression Class 10
Summary
TLDRThis educational video delves into the concept of arithmetic progression (AP), illustrating its definition through a sequence with a common difference of 3. It explains the formula for finding the nth term of an AP and applies it to various examples, including finding specific terms and proving sequences as APs. The script also covers special cases like constant sequences and negative differences, and offers strategies for solving complex problems involving APs, such as finding terms from the end and using the sum and product of terms. The video aims to make arithmetic progression easy to understand and encourages practice for mastery.
Takeaways
- 📚 The sequence 1, 4, 7, 10, 13, 16, 19 is an example of an arithmetic progression (AP) because the difference between consecutive terms is constant (3 in this case).
- 🔢 The general formula for finding the nth term in an arithmetic progression is \( T_n = a + (n - 1) \times d \), where \( a \) is the first term and \( d \) is the common difference.
- 📉 An arithmetic progression can also be decreasing, such as the sequence 5, 4, 3, 2, 1, where the common difference is -1.
- 🔄 A sequence where all terms are the same, like 3, 3, 3, ..., is also considered an arithmetic progression with a common difference of 0.
- 📝 To prove a sequence is an AP, one must show that the difference between any two consecutive terms is constant.
- 📈 The nth term formula can be used to predict terms in an AP, as demonstrated by finding the sixth term of the sequence 1, 4, 7, 10, 13, 16, 19.
- 🔑 The key to solving problems involving APs is to use the formula for the nth term and the concept that the common difference is constant between consecutive terms.
- 🧩 When dealing with problems that involve finding terms in an AP, it's helpful to set up equations based on the given conditions and solve for the unknowns.
- 📌 Remember that the formula for the nth term from the end of an AP is \( L - (n - 1) \times d \), where \( L \) is the last term.
- 📉 When calculating the number of multiples of a number within a range, such as multiples of 4 between 10 and 250, an AP approach can be used to find the count.
- 📝 For solving problems with a given number of terms in an AP, it's often more convenient to use terms like \( a - d, a, a + d \) for three terms, and \( a - 3d, a - d, a + d, a + 3d \) for four terms, to simplify calculations.
Q & A
What is an arithmetic progression (AP)?
-An arithmetic progression is a sequence of numbers in which the difference between consecutive terms is constant. This constant difference is denoted by 'D'.
How is the nth term of an arithmetic progression calculated?
-The nth term (T_n) of an arithmetic progression can be calculated using the formula: T_n = a + (n - 1) * D, where 'a' is the first term and 'D' is the common difference.
What is the difference between an arithmetic progression that is increasing and one that is decreasing?
-An increasing arithmetic progression has a positive common difference, meaning each term is larger than the previous one. A decreasing arithmetic progression has a negative common difference, indicating each term is smaller than the previous one.
Can an arithmetic progression have a common difference of zero?
-Yes, an arithmetic progression can have a common difference of zero. In such a case, all terms in the sequence are the same, as there is no change between consecutive terms.
What is the formula for finding the nth term from the end of an arithmetic progression?
-The formula for finding the nth term from the end of an arithmetic progression is: T_n = L - (n - 1) * D, where 'L' is the last term and 'D' is the common difference.
How can you determine if a sequence is an arithmetic progression without using the formula?
-You can determine if a sequence is an arithmetic progression by checking if the difference between any two consecutive terms is constant throughout the sequence.
What is the sum of the first three terms of an arithmetic progression if the common difference is 2 and the first term is 3?
-If the first term (a) is 3 and the common difference (D) is 2, the sum of the first three terms is 3 + (3 + 2) + (3 + 2*2) = 3 + 5 + 7 = 15.
How many multiples of 4 are there between 10 and 250?
-There are 60 multiples of 4 between 10 and 250. This is calculated by considering the sequence of multiples of 4 as an arithmetic progression with a first term of 12 and a common difference of 4, and solving for the number of terms (n) using the formula T_n = a + (n - 1) * D.
What is the trick to solving for the terms of an arithmetic progression when given three terms and certain conditions?
-The trick is to express the three terms as a - D, a, and a + D, which simplifies the process of solving for the common difference (D) and the first term (a) when given certain conditions.
How do you find the middle term of an arithmetic progression with a known first term, common difference, and last term?
-To find the middle term of an arithmetic progression, first determine the total number of terms (n) using the formula for the last term (T_n = a + (n - 1) * D). The middle term is then T_(n/2), which can be calculated using the formula for the nth term with the appropriate value of n.
Outlines
📚 Introduction to Arithmetic Progression
This paragraph introduces the concept of an arithmetic progression (AP), a sequence of numbers with a constant difference between consecutive terms. The script uses the sequence 1, 4, 7, 10, 13, 16, 19 to illustrate this, highlighting the constant difference of 3. The presenter encourages viewers to practice different types of questions on AP and provides a link to their website for further practice. The basic formula for finding the nth term in an AP, TN = a + (n - 1) * D, is explained, where 'a' is the first term and 'D' is the common difference. An example is given to find the sixth term, which is correctly calculated as 16.
🔍 Identifying and Simplifying APs
The script continues with an exercise to identify an arithmetic progression and find the next term. It simplifies complex terms involving square roots and demonstrates that the common difference is constant, which confirms the sequence is an AP. The next term is found by adding the common difference to the last simplified term, resulting in 9 root 2 or √162. The paragraph also discusses how APs can be decreasing and includes an example with a common difference of -1, as well as a constant sequence where the common difference is zero.
📘 Solving Problems Involving APs
This paragraph presents a series of problems related to arithmetic progressions, including finding the next term in a given sequence, determining the values of 'a' and 'b' such that a sequence forms an AP, and finding the 21st term of a specific AP. The paragraph emphasizes the importance of using the AP formula TN = a + (n - 1) * D to solve these problems. It also provides a method to solve for the first three terms of an AP when given certain conditions, using two equations derived from the AP formula.
🔢 Calculating Terms and Multiples in APs
The script discusses how to find the middle term of an AP and how to determine the number of multiples of a certain number within a specified range. It explains that if a sequence is given to be in AP, one can directly use the formula for the nth term to find the middle term without proving the sequence is an AP. An example is provided to find the 12th term in a sequence with 23 terms. Additionally, the paragraph explains how to calculate the number of multiples of 4 between 10 and 250, forming an AP and using the formula to find there are 60 such multiples.
📝 Special Cases and Formulas for APs
This paragraph introduces special cases for calculating terms in an AP from the end, using the formula TN = L - (n - 1) * D, where 'L' is the last term. It also provides a trick for solving sums using the standard AP formula and explains the importance of selecting terms in an AP for different numbers of terms, such as three, four, or five terms, to simplify calculations. The paragraph emphasizes the convenience of using specific term selections for solving problems involving the sum or product of terms in an AP.
🎯 Solving Complex AP Problems
The script presents a complex problem involving an AP with three terms, where the sum and a particular product condition are given. It demonstrates how to use the convenient form of the first three terms (a - D, a, a + D) to simplify the problem and solve for the common difference 'D'. The paragraph concludes with the solution to the problem, identifying the AP as 7, 16, 25, and so on, and invites viewers to attempt a similar problem involving dividing 32 into four AP terms with a specific product ratio.
📌 Conclusion and Further Resources
In the concluding paragraph, the presenter summarizes the video's content on arithmetic progressions and encourages viewers to apply the concepts learned. They mention that formulas for calculating the sum of AP terms will be discussed in a separate video and invite viewers to like, comment, share, and subscribe for more content. The presenter also directs viewers to their website and social media for additional quizzes, questions, and courses.
Mindmap
Keywords
💡Arithmetic Progression (AP)
💡Common Difference (D)
💡First Term (a)
💡Term Number (n)
💡nth Term Formula
💡Decreasing AP
💡Constant AP
💡Sum of Terms
💡Multiples of a Number
💡Product of Terms
💡Middle Term
Highlights
Arithmetic progression (AP) is a sequence where the difference between consecutive terms is constant.
The common difference in an AP is denoted by 'D' and the first term by 'A'.
The nth term of an AP can be found using the formula TN = A + (n - 1) * D.
An example calculation shows the 6th term of an AP is 16, using the formula.
APs can also be decreasing sequences with a negative common difference.
A sequence with all terms the same forms an AP with a common difference of zero.
The video includes practice questions to demonstrate the application of AP concepts.
An example shows how to prove a sequence is an AP by finding a constant common difference.
The formula for finding the nth term from the end of an AP is discussed.
A trick for solving sums in APs using the nth term formula is introduced.
The video explains how to find the middle term of an AP given its first and last terms.
A proof is provided to show that if M times the Mth term equals N times the Nth term in an AP, then the (M + N)th term is zero.
A special case is discussed where the sum of the first three terms of an AP is given, and a relationship between terms is used to find the AP.
The video demonstrates how to find the number of multiples of a number within a range, using an AP formed by those multiples.
A method for selecting terms in an AP for easier calculation when given a certain number of terms is explained.
The final question challenges viewers to divide a number into four AP terms with a specific product relationship.
The video concludes with a summary of the key points and an invitation to engage with the content.
Transcripts
1 4 7 10 13 16 19 friends do you know
what is this sequence of numbers known
as that's right it's called an
arithmetic progression because if you
take the difference of any two
consecutive numbers can you see that the
difference is constant it's 3 in this
case and arithmetic progression is going
to be the topic of this video in this
video we are going to be practicing
different types of questions on this
topic and I'm sure after watching this
video you'll find arithmetic progression
really easy and if you find this video
useful to hit the like button and share
it out with your friends and do check
out my website mono chai academy.com to
practice more questions on this topic
I'll put the link below
first let's revise the basics of
arithmetic progression or it's known as
AP in short so here I have this sequence
of numbers 1 4 7 10 13 16 19 and so on
now why is this in arithmetic
progression because if you take the
difference of a term and the previous
term you will see that the difference is
constant so can you see that 4 minus 1
is 3 7 minus 4 is also 3 and here if you
take 30 minus 10 3 19 minus 16 is also 3
so can you see that they have a common
difference and that common difference is
denoted by the letter D and it's 3 for
this sequence right and the first term
is denoted by the letter A so now if you
want to find the n it term in this AP
then we can express the NH term as TN or
you can use the symbol and for it and
what is the formula for that TN equals a
plus n minus 1 into D where is the first
term n is the the term number that you
want to find and D is the common
difference so let's try out this formula
let's say we want to find the sixth term
in this AP okay so we want to find T six
right and what is that going to be so
using this formula e so we'll start with
E and what is the value of e that's
right the first term can you see it's
one so let's substitute that and then
since we're interested in the sixth term
here it's going to be n is going to be
six here right so we need to substitute
6 minus 1 and what is the common
difference here as we saw the common
difference was 3 so it's going to be
multiplied by 3 and if you calculate
this what do you get it's going to work
out to be 1 plus 5 into 3 1 plus 15
which is 16 so the sixth term of this AP
is 16 and can you see one two three four
five six so the formula using the
formula we predicted the sixth term so
this formula TN or you can write it as a
n gives you the expression for finding
the N its term in the arithmetic
progression okay now let me ask you
let's say we have a numbers like this 5
4 3 2 1 okay so do you think this is an
arithmetic progression that's right
this is also an arithmetic progression
because if you take the difference of a
term and the previous one what is the
common difference here so if you take
the difference between these two terms
can you see the difference is minus 1
and again any other two terms you take
the difference is minus 1 so can you see
that the common difference here is the
same it's minus 1 so this is also an AP
so an AP can be decreasing also like
this and how about this equals let's say
I write this sequence here 3 comma 3
comma 3 comma 3 comma 3 and so on
so is this an AP what do you think
that's right
this is also an AP again apply the same
concept can you see that the common
difference between two consecutive terms
here is zero right so the common
difference can you see here is zero so
even this cons the same terms 3 comma 3
comma 3 comma 3 also form an arithmetic
progression now let's practice some
questions on arithmetic progressions I
would recommend you to pause the video
here and go and get a pen and paper so
that we can solve the questions together
so are you ready let's start with our
first question show that the progression
below is an AP find the next term of the
AP and here's the progression that we've
been given now how do we solve this so
the trick is to show that this
progression is in AP we need to find the
common difference between the terms and
we need to show that this common
difference is constant right because
remember in an AP the common difference
should be constant but since the terms
look a little complicated here since
they're square root 18 square root 50
first let's see if we can simplify each
term so we can write square root 18 as
so if we take its 9 into 2 right so it's
going to work out to be 3 root 2 ok and
what will this term work out to be so
try it it's going to be 5 root 2 right
and this term is going to be so 98 is 49
into 2 so that's if you take the square
root of 49 so we have 7 root 2 right so
now the terms are simplified so let's
find the common difference between
consecutive terms so between this term
and its previous term so what is the
difference going to work out to be so if
you subtract these two terms what do you
get right so if you subtract it it's
going to be
- root - right so if you subtract these
two similarly if you subtract these two
what are you gonna get it's a hand going
to be 2 root 2 right 7 root 2 - 5 root 2
so can you see that the common
difference is 2 root 2 so therefore we
have proved that this is an arithmetic
progression this series is in AP ok and
now we need to find the next term of the
AP so what term will come after this so
that's pretty simple to find that term
you just need to add the common
difference to this term right so it's
going to be 7 root 2 plus the common
difference D which is 2 root 2 so what
is that term going to be
the next term is going to be 9 root 2
okay and if you want to write it in this
form so we need to take the 9 inside so
that's going to be 9 into 9 into 2 so
the answer is going to be once root of
162 right so that's the next term of
this arithmetic progression that's
around say no now let's try this
question find a and B such that the
numbers a 9 b + 25 form an AP so again
what's the key concept to use here that
since these numbers form in AP the
difference between a term and its
previous term is going to be constant so
let's apply that here so the difference
between 9 and a is going to be 9 minus a
right and that's going to be equal to
the difference between B and 9 so B
minus 9 right and that's going to be
equal to 25 minus B because we know that
the difference is constant it's a common
difference okay so now let's use this
part and so
for B so we have B minus 9 equals 25
minus B right so if you solve that it's
going to be 2 B equals 34 so B works out
to be 34 by 2 which is 17 so we found
the value of B now we can use this part
9 minus a equals B minus 9 to easily
find it so let me write that down here 9
minus a equals B minus 9 right
I'm using this part and since we already
have the value of B so it's going to be
9 minus a equals 17 minus 9 so 9 minus a
is equal to 8 and so basically we have
if you solve it is equal to 1 so there
we found our answer we've got the value
of a and the value of B and we use this
important thing that the difference
between two consecutive terms is
constant so 1 9 17 and 25 form an AP
here can you see that the common
difference is 8 ready for the next
question
here it is find the 21st term of the AP
minus 5 minus 5 by 2 0 5 by 2 and so on
okay and remember to find the 21st term
or the NH term we need to use the
formula TN equals a plus n minus 1 into
D remember that formula okay and so what
is the first term here it's denoted by a
right and a is going to be this term
minus 5
now what is the common difference D now
since we know already know that this is
an AP we don't need to prove it's an AP
so we can take the difference of any two
terms to find the common difference D
and let's see what do we get so it's
easy to take these two terms five by two
minus zero so the common difference is
basically five by two right so we've got
our D the common difference here okay
and since we need to find the twenty
first term so we need to find T 21 or
you can also denote it by a 21 okay so
that's going to be a plus so we're going
to substitute the value minus five here
plus n so here n is going to be twenty
one minus one times the difference five
by two okay and if you work that out so
this is twenty one minus one that's
going to be 20 into 5 by 2 so we are
gonna get here minus 5 Plus this is
going to be 50 this works out to 50 so
our twenty first term is 45 that is our
answer here so here the main thing is to
use this formula and correctly
substitute the value of a n and D to
solve this question now let's look at
this question some of the fourth and
eighth terms of an AP is 24 and the sum
of its 6th and 10th terms is 44 find the
first three terms of the AP so the trick
to solving this is you look at the given
conditions and you form your equations
okay so what are the conditions that
we've been given that the sum of the
fourth term so T 4 plus T 8 the eighth
term is given to us as 24 okay so that's
our first condition and now let's look
at the other condition we've been given
that the sum of the sixth term so T 6
plus the 10th term t10 is given to us as
44 okay
so let's use our arithmetic progression
formula for the terms and let's see what
equations we can get from here so let's
start with the first one here T 4 plus T
8 is 24 now we are going to use the
formula remember T N equals a plus n
minus 1 into D okay so if you apply this
formula T 4 is going to be a plus 3 D
right so we are doing it for this
equation plus we are going to get a plus
70 and that's going to be equal to 24
right and so if you simplify that we get
2 A plus 10d is 24 and further
simplifying that we get a plus 5 D is
equal to 12
okay so that's our first equation here
ok and now let's use the second
condition this one and let's form the
other equation so it's going to be a
plus 5 D right so that is T 6 and T 10
is going to be a plus 90 right and
that's given to us as 44 so let's add
these up we get 2 A plus 14 D is 44 so
therefore you can write this as the
simplified form will be a plus 7 D is 22
so this is our second equation here
right and can you see that we have two
linear equations here one and two and we
have two unknowns two variables right E
and D so clearly you can see we have two
linear equations in two variables so you
can solve them pretty easily right so
I'm not going to show the solving of
linear
equation so you can do that it's pretty
easy so if you solve them we are going
to get a as minus 13 and D is going to
be fine the common difference D so on
solving these two equations equation 1
and 2 we get these values and so now
it's pretty easy to find the first three
terms of the AP so what will be the
first three terms so the first term
obviously is going to be this one the
first term is minus 13 simply a right
the first term the second term is going
to be a plus D right T 1 sorry T 2 right
so that's a plus D so that's going to be
minus 13 plus 5 so we have minus 8 and
the third term is going to be a plus 2 D
correct and so what is that going to be
so we are going to get minus 3 so here
we have solved the first three terms of
the AP minus 13 minus 8 and minus 3
let's try this question find the middle
term of the AP - 13 to 0-5 197 and all
the way up to 37 okay so this sequence
that we've been given the question says
that it's in AP now if it's not given
that it's an AP then you first need to
prove that it's an AP by checking the
difference should be constant between
the terms so we don't need to do that
right we can assume that it's in AP
because it's been given so what do we
know about this AP we know that the
first term is 213 so this is a right so
is 213 okay and what is the difference
the common difference here so it's going
to be 2 0 5 - 2 13 right so that's going
to be minus 8 okay and we need to find
the middle term but we don't know how
many terms this AP has right so first we
need to find the number of terms in the
AP okay so let's say the last term this
is the NH term of the AP
okay so let's say this is the and it's
dumb so we can represent TN the Anette
term we know is 37 okay and what is the
formula for the NH term it's going to be
a plus n minus 1 into D right so now if
we substitute these values so 37 is
going to be to 13 plus n minus 1 into
the difference minus 8 and so this is 37
right and if you solve this
you'll get n the number of terms of the
aps 23 right so what do you think is
going to be the middle term of this ap
since the AP has 23 terms the middle
term is going to be so to find the
middle term that's going to be 23 plus 1
by 2 right so the 12th term is the
middle term right and so how do we take
out find the middle term here so we
basically need to solve for T 12 and
that's again a plus n minus 1 into D so
that's going to be 2 13 and n minus 1 is
going to be 11 into D the difference is
minus 8 okay and if you solve this you
will get T 12s
as 125 so that's our final answer
we found the middle term t12 is 125 now
let's look at this question if M times
the mhm term of an AP is equal to M
times the NH term and M is not equal to
n show that it's M plus n it term is
zero okay so we need to show that the M
plus n eighth term is zero
so first let's use the formula and write
the terms that we are interested in so
the MS term we can express it as TN is
equal to a plus M minus 1 right into D
and what will the NH term be same thing
just use that formula TN is going to be
equal to a plus n minus 1 right n minus
1 into D okay and we are interested in
the M plus n its term okay so let's
write that down also so we know what
term you are interested in so T M plus M
is equal to a plus M plus n minus 1
right into D okay so that's the term we
are interested in now let's see what
have we been given so we can form the
equation on what's what's given to us in
the question so we've been given that M
times the MS term is equal to n times
the NS Tom so let's write that down in
symbol form so we've been given M times
the MS term right is equal to n times
the NH term okay so let's go ahead and
substitute these formulas in this
equation
so we'll get M times a + M minus 1 into
D is equal to n times a plus n minus 1
into D right okay and let's multiply
that out and rearrange that so we'll get
ma plus M into M minus 1 times D and
that's equal to ne plus n into n minus 1
into D okay that's pretty simple
and now let's group the a and D terms
together so we can write into M minus n
so I'm bringing the N a that side plus
we will get M squared minus M minus N
squared plus n okay into D and that's
equal to 0 so I brought all the terms on
this side and we've grouped them
together so now let's see what we can do
here so can you see that we have a a
into M minus n here plus you can group
these together so M square minus n
square so that's going to be that's
essentially M plus n into M minus n
right and then we have this remaining
term here minus of M minus M into D
equals 0 okay so we are just simply
grouping the terms together and seeing
what we can get in common here so now
can you see that the M minus n is in
common right so let's take that out here
and we are going to get a plus and what
are we left with if you take M minus n
common it's going to be M plus n minus 1
into D equals 0 right so just take a
look at this so you can compare this
equation the one we have here right just
by grouping this will get this and we've
been given in the question that M is not
equal to n right so this M minus n term
basically this term is not 0 so this is
not equal to 0 ok we can take that to
the right-hand side and so so we are
going to be left with the term a plus M
plus n minus 1 times D is equal to 0
right because this term is 0 this term
is not 0 and what is this expression
here a plus M plus n minus 1 times D so
if you carefully look at this one and
compare it it's exactly R T M plus n
right because T M plus n is a plus M
plus n minus 1 into D so therefore we
know that T M plus 1 is equal to 0 and
that's exactly what we had to prove in
this question so the key thing is to
solve these type of questions just use
the formula and write the term in terms
of TM TN and what you are interested in
right T M + n so we go on the lookout
for that term and we got it finally here
a plus M plus n minus 1 times D equal to
0 and so TM plus n is 0 till now for
arithmetic progressions we've been using
the formula of the NH term as TN equals
e plus n minus 1 into D note the NH term
here is when you count n from the
beginning right the terms are counted
from the beginning of the arithmetic
progression
but let's say you want the NH term from
the end so not from the start we will
count the term numbers from the end and
we are interested in the NH term from
the end then the formula becomes L minus
n minus 1 into D ok where L is the last
term of the AP and D is again the common
difference and n is the NH term that
you're interested in
now an easy way to remember this formula
is you compare the two formulas ok so
all you have to do is replace the first
term a with the last term L right
because we are counting from the end and
you just have to change this plus into a
- ok the rest of the part remains same n
minus 1 into D so remember that easy we
just changed the a to L and the plus to
a minus and you'll get the formula for
the N its term from the end I'm also
going to be showing you a special trick
where you can solve the sums just with
this TN formula and you don't need to
use the NH term from the end formula so
you can solve it just using the first
formula it's coming up are you ready for
the next question here it is find the
eighth term from the end of the AP 7 10
13 all the way up to 184 so since we
need to find the eighth term from the
end let's see if we can use the formula
of the NH term from the end that we just
learnt okay so let's write down that
formula which was the NH term from the
end right
and what is that formula the NH term
from the end is L the last term -
remember there will be a minus there and
n minus 1 times D ok so what is the last
term in this AP so that's pretty simple
the last term is 184 so let's substitute
that here 184 is our last term minus n
minus 1 now we want the eighth term so n
is going to be 8 so 8 minus 1 times the
common difference and what is the common
difference here so you subtract these
two terms 10 minus 7
the common difference is 3 right and so
if we calculate that that's going to be
184 minus 21
so the NH term that is sorry the eighth
term from the end is going to be 163 so
that's our answer the eighth term from
the end the eighth term from the end is
163 okay now let me show you a special
trick how to solve this question without
using this formula we are going to use
our normal formula the TN equal to a
plus n minus 1 into T so we'll use the
formula of the nth term from the
beginning so the trick of doing that is
just take the series the AP that you've
been given and just reverse it okay so
let's write the series in Reverse here
so it's going to be 184 comma 13 comma
10 and 7 right now why are we reversing
the series because we want to use the
formula the normal formula of the NH
term from the beginning right so in that
formula was a plus n minus 1 times D ok
and what is our first term here
we have reversed the series this is the
first on right a so first term is going
to be 184 and we want to find the eighth
term from the end okay so for the
reverse series it's gonna be the eighth
term from the beginning right if you
think about it the eighth term from the
end of the AP was what we wanted to find
now since we reverse the series it's
going to be the eighth term from the
beginning so we can easily apply this
formula and n is going to be eight so we
are basically calculating te8 right the
eighth term from the beginning and what
is the common difference here so the
common difference we need to subtract
the terms right so it's going to be 7
minus 10 not 10 minus 7 right you always
take the the next this term minus the
previous term so the common difference
is going to be minus 3 here and if you
solve that can you see that are you
getting the same thing so 184 plus and
this is going to be a minus 21 so it's
basically 184 minus 21 and the eighth
term from the beginning of this series
is 163 so can you see we got exactly the
same answer no matter which method we
applied so either if you are comfortable
with this formula and its term from the
end you can directly use it L minus n
minus 1 into D so be careful it's the
last term and it has a minus sign this
formula has a plus sign and then you
substitute it and you'll get the eights
term from the end or if you want to use
this technique you just reverse the
series and then it's a normal AP
question right you use the first term
and you just substitute n from the
beginning and just take care to
calculate the right common difference so
we can solve it just from the basic
formula that we know I have an
interesting question for you how many
multiples of 4 lie between 10
two-fifty okay so what do we need to
find here so between 10 and 250 so how
many multiples are the of four are there
between 10 and 250 right so between
these two numbers
now one important thing to note here is
since they've asked between 10 and 250
we don't include the numbers 10 and 250
it's the numbers in between but if it
was from 10 to 250 then you include the
numbers okay
so remember that in between 10 and 250
these two numbers are excluded we don't
include them but if the question says
from 10 to 250 then you need to include
these numbers also okay so how many
multiples of 4 we need to find so what
are the multiples of 4 we know that 10
is not a multiple so the first multiple
of 4 is 12 here 16 20 right and it's
going to be all the way up to will it be
up to 250 no because 250 is not a
multiple of 4 so what is the last
multiple of 4 here 248 and even if 250
was a multiple of 4 since we need to
find the numbers in between we would
have excluded it so basically the
numbers we are interested in are from 12
to 248 and these are the multiples of 4
so now can you see that these form an AP
so we have an AP here right because as
you can see the common difference is 4
there are multiples of 4 so now what is
the first term of this AP as you can see
the first term is 12 so we have a equal
to 12 right and what is the common
difference simple 16 minus 12 the common
difference as expected is 4 for the
multiples of 4 right and then this is
our NH term in the AP right so our last
term here is TN is 248 and we need to
find
how many multiples so what do we need to
find here we need to find em how many
multiples are there so let's go ahead
and use our formula TN equal to a plus n
minus 1 times D right and so if we
substitute TN is 248 equals 12 plus n
minus 1 so we need to find n that's our
variable here and D is 4
okay so you basically need to solve this
equation for N and if you solve it you
get NS 60 so that's our answer that
there are 60 multiples of 4 lying
between 10 and 250 now let's talk about
an important point on how to select the
terms in an AP for example let's say
you're given that an AP has three terms
so what do you express the three terms
as you might be thinking a a plus D and
a plus 2 D but let's talk about what's a
more convenient way of expressing three
terms in an AP okay so you should
actually use a minus D a and a plus D
instead of a a plus DN a plus 2 D it's
easier to use these terms a minus T E
and a Plus D and can you see what is the
common difference here that's right the
common difference is D but let's say
you're given that four numbers are in AP
then what do we take the four terms as
so when four numbers are in AP we need
to use a - 3 D a - d a + D + a + 3 D
here okay and what is the common
difference in this case can you see that
the common difference is 2 D here okay
because if you subtract this term and
the preceding one the previous term you
will get a difference of 2 D and if
you're given that five numbers are in AP
then what can we take
then we are going to take the terms as a
- 2 d e - d e a + D and the fifth term
is a plus 2 D now here the common
difference again is as you can see D
okay so remember for the three terms the
common difference was D for the four
terms it's to D okay and for the five
terms its D so you can easily remember
that because four is an even number so
the difference is to D an even number
right and these ones are odd number of
terms three and five so their common
difference is just D now let's look at
an example to see why these type of
terms are more convenient than taking a
a Plus D a a plus 2 D so why should we
take the terms as this let's take a look
at the example here's our next question
the sum of first 3 terms of an AP is 48
if the product of first and second terms
exceeds four times the third term by 12
find the AP okay so how do we solve this
question we've been given that the sum
of the first three terms of an AP is 48
and remember as we learnt what is the
convenient terms that we can take here
for three terms so rather than taking a
a plus T and a plus 2 D we're going to
take these three as our first terms a
minus T okay a and a plus T
so let these be the first three terms of
the AP and since it says the sum of the
terms is 48 so let's see what will the
sum of the three terms be so this Plus
this Plus this right so it's going to be
a minus T plus a and plus a plus D and
the sum is given as 48 now here you can
see the D will cancel out right and so
we will have 3a equals 48 and so what is
the value of a it's 16 so can you see
how convenient it was because we took
these three as the first three terms the
D got cancelled out we took a minus D a
and a plus D and so it is very simple to
solve for a okay and now let's form the
equation based on what's what's the
other condition given to us so it says
that if the product of the first and
second terms exceeds four times the
third term by 12 so let's write that in
symbol form here so it's going to be the
product of first and second term so
that's t1 and times t2 exceeds four
times the third term three t3 so 43 by
12 so that's our other equation again
let's substitute the terms so this is
our first term second term and third
term right
so that's t1 t2 t3 so we can substitute
T one is going to be a minus D times a
minus four times a plus T equals 12 and
we already know the value of a so let's
use that
so we'll get 16 minus D times 16 minus
four
into 16 plus D equals 12 and if you
solve this equation
you'll get the value of D and that's
going to turn out to be 9 okay so here
we've solved for E and D so it's very
easy to find our AP now right so the
first term of the AP is going to be e
minus D 16 minus 9 so our AP is going to
turn out to be 16 minus 9 that's 7 and
then you have the second term is a so
that's 16 and the third term is e plus T
16 plus 9 25 and so on so we've found
our AP by solving these equations based
on the conditions that were given to us
so the main trick is you look at the
question and form the equations based on
the condition and here we use the trick
that the first three terms are given so
rather than using a a plus D and a plus
2 D we use this convenient form a minus
D a and a plus D and that made it really
easy to solve this question and so our
AP turned out to be 7 16 25 and so on
and here's our final question divide 32
into four parts which are the four terms
of an AP such that the product of the
first and four terms is to the product
of the second and the third terms as
seven is to 15 I want you to try this
question yourself and do let me know
your answer by putting it in the
comments below I look forward to reading
your comments so friends I hope the
concept of arithmetic progression is
crystal clear to you now
there are also formulas to calculate the
sum of the terms of an AP but we'll
discuss that in a separate video and do
remember to like comment and share out
this video with your friends and if you
haven't subscribed to my youtube channel
already hit the subscribe button
now also click on the notification bell
to get notified about new videos you can
check my Facebook page and do check out
my website Manoj academy.com for the
quiz and the top three questions on this
video and for more courses and videos
for you I'll put the links below thanks
for watching
Ver Más Videos Relacionados
Matematika SMA - Barisan dan Deret (1) - Barisan Aritmatika, Rumus Barisan Aritmatika (A)
Harmonic Sequence (Tagalog/Filipino Math)
Arithmetic Sequences and Arithmetic Series - Basic Introduction
Harmonic Sequence
PROBLEM SOLVING WITH PATTERNS || MATHEMATICS IN THE MODERN WORLD
Grade 10 MATH Q1 Ep3: Write and Use the Formula of the nth Term of an Arithmetic Sequences
5.0 / 5 (0 votes)