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