Arithmetic Progression Class 10

Summary

This 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

00:00

📚 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.

05:03

🔍 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.

10:08

📘 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.

15:09

🔢 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.

20:12

📝 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.

25:13

🎯 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.

30:16

📌 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.

Keywords

💡Arithmetic Progression (AP)

Arithmetic Progression (AP) is a sequence of numbers in which the difference between the consecutive terms is constant. In the context of the video, the AP is the central theme, with the sequence 1, 4, 7, 10, 13, 16, 19 being used as a primary example to illustrate the concept. The video explains how to identify an AP by looking for a constant difference between terms and how to calculate specific terms within the sequence using the formula TN = a + (n - 1) * D, where 'a' is the first term, 'n' is the term number, and 'D' is the common difference.

💡Common Difference (D)

The common difference (D) in an arithmetic progression is the constant value that you subtract from one term to get the next. It is a fundamental aspect of defining an AP. In the script, the common difference is used to demonstrate how to identify an AP and to calculate the nth term of the sequence. For example, in the sequence 1, 4, 7, 10, the common difference is 3, as each term increases by 3 from the previous one.

💡First Term (a)

The first term (a) of an arithmetic progression is the initial value in the sequence. It serves as the starting point for calculating subsequent terms. In the video, the first term is used in the formula to find any term in the AP, and it is denoted by 'a' in the formula TN = a + (n - 1) * D. For instance, in the sequence 1, 4, 7, ..., the first term 'a' is 1.

💡Term Number (n)

Term number (n) refers to the position of a term within an arithmetic progression. It is used in the formula to calculate the value of any term in the sequence. The video script uses term number to explain how to find specific terms, such as the sixth term or the 21st term, by substituting 'n' into the formula TN = a + (n - 1) * D.

💡nth Term Formula

The nth term formula is used to calculate the value of any term in an arithmetic progression given the first term, the common difference, and the term number. The formula presented in the video is TN = a + (n - 1) * D. It is central to solving various problems related to APs, such as finding a specific term or verifying the progression's nature by checking the difference between terms.

💡Decreasing AP

A decreasing AP is an arithmetic progression where the common difference is negative, meaning each term is smaller than the previous one. In the script, the sequence 5, 4, 3, 2, 1 is given as an example of a decreasing AP, where the common difference is -1, demonstrating that APs can also decrease in value.

💡Constant AP

A constant AP is a special case where all terms in the sequence are the same, and the common difference is zero. The script mentions the sequence 3, 3, 3, ... as an example of a constant AP, showing that even when the terms do not change, the sequence can still be considered an arithmetic progression.

💡Sum of Terms

The sum of terms in an arithmetic progression refers to the total of all the numbers in a given sequence. While the video does not delve into the formula for the sum of an AP, it does mention the sum of the first three terms of an AP as part of a problem-solving example, emphasizing the importance of understanding how terms relate to each other in an AP.

💡Multiples of a Number

Multiples of a number are the product of that number and an integer. In the context of the video, the concept is used to find how many multiples of 4 lie between 10 and 250, which forms an arithmetic progression with a common difference of 4. The video uses this concept to illustrate another application of APs in solving real-world problems.

💡Product of Terms

The product of terms refers to the result of multiplying several numbers together. In the script, the product of the first and fourth terms of an AP is compared to the product of the second and third terms, as part of a problem that requires understanding the relationship between different terms in an AP to find the specific sequence that satisfies the given condition.

💡Middle Term

The middle term of an arithmetic progression, when the number of terms is odd, is the term that appears exactly in the center of the sequence. The video explains how to find the middle term of an AP with an odd number of terms by using the formula for the nth term and recognizing that the middle term is at position (n+1)/2, where 'n' is the total number of terms.

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