Arithmetic Progression (AP) is a sequence where each term increases by a constant difference 'd'. The key formulas are: nth term a_n = a + (n-1)d, sum of n terms S_n = n/2[2a + (n-1)d], and alternative sum formula S_n = n/2(a + l) when the last term 'l' is known. To solve AP problems, identify the first term 'a', common difference 'd', and number of terms 'n', then apply the appropriate formula. For complex problems, set up equations using the nth term and sum formulas, solve the system to find unknown parameters, and verify solutions by checking if the calculated values satisfy all given conditions.
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Master Arithmetic progression in 10 Mins! | Class 10th Maths MagicAdded:
Today I'm going to deal with sum related. So in arithmetic provision as I told to everyone we are having two different formulas. So one is related to term and other is related to sum. If once you are clear with term related sums you can start out with some related.
So let's start with so let's start with the sums related to sum. So recall back as long ago we started with this chapter. So once recall back so we got few days gap in between.
So here first some uh first formula is related to term which is a n is equal to a + n -1 d. So it's related to term and then next formula we are having related to sum for sum formula is s n is = nx 2 2 a + n -1 d.
So this is related to sum.
Sometimes if the last term is given, if last term or a n is given, we use direct formula of the sum that is sn is nx 2 a + l. Here l represents last term. l is representing last term. So totally in this chapter we are having only three formulas. So as I told chapter is based on three formulas.
So today we are going to u start out the sums related to sum formula. So here uh let's start with first question. Find the sum of first 22 terms of a. A is consisting of three.
So here we need to find sum of 22 terms. We need to find sum of 22 terms of a and a is given as follow 8 3 - 2 so on we need to find sum of the terms.
So here as per the question first term a is 8 common difference B a2 minus a1 so here a2 is 3 3 - 8 so it's - 5 and n is 22. We need to find sum of 22 terms. So n is 22 here.
So let's start out with the formula s 22. Some formula is sn is n /x 2 2 a + n -1 d.
So here n is given 22. So it is 22x 2. 2 a is 8 + 22 -1 common difference is given here - 5. So let's restart it. Today we started with sum formula. So here sum of 22 terms of a we need to calculate. Terms are given as follows 8 3 - 2 so on. So from the question first value first number is first term.
So a is 8. Now common difference d can be calculated by subtracting a2 minus a1. So here the terms as follows a1 a2 a3 so on. So common difference is a2 minus a1 a2 is 3 a1 is 8. So 3 - 8 is - 5. We got the value of a. We got the value of b and terms are 22. We need to find sum of 22 terms. So formula of sum is sn is equal to n /x 2 2 a + n minus1 d. So here n value is 22.
Replacing n with 22 22 by 2 2 a is 8. So replacing a with 8 as a is given 8 and n is 22 so it is 22 - 1 common difference is -5 so let's calculate now so here 11 uh I'm not getting your question clearly ma'am is the power of x by the derivative of so your question is incomplete.
Okay. So now 11 28 is 16.
Now 22 - 1 is 21 * - 5. So it's 11 16 + -.
So final answer 11 * now plus - we need to subtract it 11 * 89.
So those who are attending once check out the calculation whether it is correct or not. So this is the final answer. We got the sum of 22 terms as 979.
So as bigger number is negative here. So - 89. So final answer is - 979.
So some related questions are very simple. They are purely based on formula. So if you're clear with formula, you can easily solve some related sums. So let's start with one more question here. Find the sum of Okay, how many terms are taken? So here we need to find sum of first,000 positive integers.
First, thousand positive integers.
So here first, thousand numbers are there. So from this we can clarify that n is 1,000 year. So question uh I'm dealing with 10 standard.
So 10th standard chapter number uh five arithmetic progression is going on.
So now sum of first,000 terms. So n is given 1,000 and here terms are not given positive terms are given. So we need to write the sequence here. So sequence will be 1 2 as positive terms are there. So sequence will be 1 2 3 4 so on. So from this given data we can calculate the different values. So a is the first number. Thus a is 1. Common difference is a2 minus a1. So it's 2 minus 1 is 1.
Now we need to find sum. So let's put out all together in sum formula. So sn is n /x 2 2 a + n -1 b. Now here n is 1,000. So it's 1,000 upon 2 after a long result.
So it's 2 a is 1 + 1,000 - 1 common difference is 1.
So 500 * by 2 1 are 2,000 - 1 is 999.
So it's 500 * 1,1. So let's calculate complete answer now.
So it's final calculation 5 1 are 5 0 0 5 1 are five. So two more zeros. So this is final answer of sum of first thousand terms.
O line.
Hello.
Are you sure?
Are you still there?
Jani I started with the questions uh sum based questions here say the formula for the sum once again I'm repeating for you Janvi are you there now Janvi excellent enter.
Okay, see here so already we did these formulas once again just to recall a n when we are calculating term so it's a + n -1 d while calculating sum it's n /x 2 2 a + n -1 d so in today's session we are dealing with sum uh or Another formula is there for sum.
It's n /x2 a + l. So sum of first question find the sum of 22 terms of a.
Sequence is given 8 3 - 2 so on. So from this data first number is a very nice idea.
10th already teachers.
It's very good.
So it's -5 and n we need to find sum of 22 terms. So the formula for sum is sn is n /x 2 2 a + n -1 d. Now here n is 22 by 2 2 a 8 + 22 - 1 * common difference is minus 5. Next is skipping 11 as it is idea. So just to proceed all the best 11 16 22 - 1 is 21 * - 5. So 21 * - 5 is - 105. Now plus - subtracting both the numbers. 105 - 16 is - 89. - 89 * 11 is - 979. So purely based on formula only.
So if you're clear with the formula simply you can solve out questions. Now here next question um how many terms of AP must be taken so that the sum is 78.
Next question how many terms of AP a taken to make sum of 78. So here sum is given 78 and sequence is given 24 21 18. So these two points are important to us. So from the sequence we can get the value of a from which book we are doing.
Uh it's uh from NCERT textbook.
Jan it doesn't matter from which book you are doing. you just uh get clear with the concept. Okay.
And after only I asked you about the fix.
So here sequence is given from the sequence first term is 24 common difference 21 minus 24 a2 - a1.
So - 3 sum is given 78. So sn is 78. In the question it's asked how many terms how many terms representing n is question mark. So n we need to calculate here. A is given first term 24. How we calculate common difference? By subtracting a2 minus a1. Second term minus first term.
So it's 21 minus 24 minus 3. Then sum is given 78. So sn is 78. We need to find the value of n. So let's put them in formula. Now formula of sum is sn is equal to n / 2 2 a + n -1 d. Now here sum is given 78 is equal to n we need to calculate as n is question mark. Then 2 a is 24 plus again n is unknown and then common difference solution easy. So multiplying 2 with 78 156 is equal to n * 24 * 2 is 48.
Now 3 * n so it's - 3 n - + 3.
Now n 48 + 3 48 + 3 is 51 - 3 n is equal to 156.
Now open out the bracket by multiplying with n. So you 51 n - 3 * n is 3 n². Now shifting all together towards one of the like towards left side.
to 3 n² minus plus change + 51 will change as - 51 n then already 156 is there so it's plus 156 now we need to find out the factors such that we'll be getting 51 by middle term factorization we are going to calculate n now so 156 And one more three is there.
Now 4 3's are 12 12 3's are 36 + 13 not getting 51. So let's retry it.
13's are 39 and uh 12 39 and 12.
So here we got 51 39 and 12. Okay. So the factors are 3 n² - 39 n - 12n + 156 is equal to zero. So simple middle term factorization those are clear with we need to be clear with the uh factorization topic also here. So sometimes middle term factorization we need to solve to find out value of n. So now here in both the terms 3 n is common. So taking three in common three common nus 3 are 39. So 13 repeat= 0.
Final factors n -3 and 3 n - 12 is = 0.
So n is either 13 or n is it will be shifting 12 that side 12 upon 3.
So 3 4 12 positive value n correct value final answer.
So here the sequence was going on decreasing and that to 33 difference is there N4 sum of four terms N four consider sum of four terms you sum of four terms 78.
Okay terms terms will be like go on decreasing 24 18 then again three will decrease. So you 158 cross what is the sum of terms and what is the sum of four terms sum of four terms check s4 is equal to so once again solving with the formula n /x 2 2 a + n -1 1 D. So n is 4 4x 2 a is 24 as per the given question about n 4 - 1 or common difference - 3 you 2 are 4 2 multiply to 48 4 - 1 is 3 3 are 9 - 9 you get 2 * 39 9.
So answer is 78.
So n cannot be 13. So n is four.
Positive value. We need to solve both the condition and then represent the final answer. So n cannot be 13. So value of n is four.
Is it clear? Should I start next question?
Jan, is it clear?
Janvi, are you there?
Okay.
Now, uh I'll be asking few questions to you. You need to uh start giving answers to me. So, here a is given. Okay.
Directly you can solve it. So here so let's see the new pattern question here.
So this is uh those who are uh having NCET textbook third question first one we are solving here a is given five b is given 3 and uh a n is given 50. We need to find n and sn. Now next find n and sn. So how to solve this kind of questions? So from the value of a n we are going to calculate n first. So this is the first step. Use the value uh formula of a n. Formula of a n is a + n -1 b a n value given here. So we need to use the formula a n a n formula a + n -1 b to you a five and we need to calculate common difference already is given value 50 questions.
towards 50. So 50 - 5 is equal to nus1 * 3. Now 50 - 5 is 45.
45 is equal to n -1 * 3.
Now n -1 is equal to 45 upon 3. So this 3 will be in the denominator of 45.
So it's 15. n - 1 is 15. So value of n is 15 + 1 16. We got the value of n. Now after getting the value of n, we need to find sum. So if this kind of questions are there there is a direct formula for sum. So here we can use sum as sn is equal to n by 2 a + l where l is nothing but a n we got 16. So replacing n with 16 16 by 2 a is 5 and l is last term which is a n 50.
So here 2 8 are 16 8 * 55.
So final answer is 440.
Is it clear?
Is it clear?
Shall I start next question if it is clear? Okay. Now, next question.
Here it is given a 7 then a13 is given 35.
A13 is given 35. What is the value of n?
A13 is given 35. What is the value of n?
Say it quickly. What is the value of n?
We need to find common difference and s3 n is 13. Very good beta. So here a n is given a3 is given 35. So we are going to calculate the value of d from there. So a3 is equal to a + n -1 b. Here n is nothing but 13. a is given 7 + 13 - 1 common difference and a 13 is given 35.
So for this kind of questions as I told we'll be shifting the number first. So it's 35 - 7 is equal to 13 - 1 is 12. So it's 12d 35 - 7 is 28 is equal to 12d.
So d is 28 upon 12. So it's 4 3's are 12 4 7s are 28. So value of d is 7 by 3.
So this we calculated the first part.
Now we need to find sum s13.
Okay. Sum of 13 terms. So whenever the last term is given as I told we need uh we can solve the question of this kind by using direct formula. Then s13 formula will be nx2 a + l. So for this kind of question if you know last term use the shortcut formula nx 2 a + l where l is the last term and last term is here 35. We need to find 13 terms sum. So n is 13 here 13 by 2 a is given 7 and l is 35.
Now adding them together 13 by 2 * 35 + 7 is 42.
So final answer is 13 1 are 13 13 2 are 26 + 127.
So sum of 13 terms is 273.
So this is final answer.
Is it clear?
Now the same kind of questions few more we are solving so that you'll get the concept clearly. Okay.
Now the next question here uh third question it's given a 12 is 37.
So first condition is given a 12 is 37 b is given 3. We need to find a and s12. So once again the same kind of question. A12 is given 37. So we are going to use the formula a n here. Now a n formula a 12 a + n -1 b a12 is given 37. So replace a12 with 37 is equal to a + n is given 12. a 12 is 37. So n is 12. 12 - 1 common difference is given 3.
12 - 1 is 11. 11 3's are 33. So 37 is equal to a + 33.
Therefore a is 37 - 33.
So it's 37 - 33. So value of a is 4. We got value of a. Now we need to find sum.
So sum formula it's going to be the simple one s12 is n by 2 a + l. So here n is 12 and value of a just now we calculated 4 and last term is given 37.
Now adding them 6 * 41.
What is L? L is A N.
A N is equal to L. L stands for last term.
So instead of L, you can write A N there if you're getting confused.
So final answer is 246.
So we can change the formula a little bit. S N is equal to N / 2 A + A N or L.
So whatever you feel easy or just follow out your textbook in your textbook whatever letter they are using. If they're using L use it otherwise A N.
Now one more question.
Next question it's given A3 is 15.
A3 is 15 and S10 is given 125.
If any another chapter started in your school or you want me to explain send that ps to me.
We need to find D and uh A 10.
See a question is different here. They started with the value of A3 and sum both are different and they're asking for D and A10 value. So from this we can conclude that we need to prepare two different equations here. So first is given A3. So a3 formula will be a + n -1 d. If you're clear enough directly you can use here 3 - 1 2 or just put out the formula. So as per your convenience. So here a + n is 3 3 - 1 common difference is not given is equal to 15. a3 is given 15. So a + 2d is equal to 15. This is first equation. Keep it aside. Now sum of 10 terms is given. So s 10 is given 125. Sum formula n /x 2 2 a + n -1 d is equal to 125 and n is 10 here. So replacing n with 10 10 by 2 2 a + 10 - 1 d is equal to 125.
Now let's have cancellation 2 5s are 10.
Even we can cancel these two also. We can cancel out if the calculations are comparison. In comparison if equal sign is there we can cancel them. So here we are canceling 125 x 5 because it is 25.
Now equation is 2 a + 10 - 1 is 9. 2 a + 9 is equal to 25. So this is second equation. We got both the equations equation 1 and two. Now let's calculate them. Let's solve them. A + 2d is = 15.
2 a + 9 is = 25.
Now multiplying the first equation with two to make them equal.
So this I'm solving by elimination method. So 2 * a is 2 a + 2 2 are 4 d is equal to 15 2 are 30. And keeping the second equation as it is now changing sign.
So this 2 a got cancel 4 - 9 is 5. So - 5d is equal to 5. Thus the value of d is -1.
We got the value of d as -1. So this value we need to put in any of the equation to find a.
So here putting the value of d in equation one. Equation one was a + 2d is equal to 15. So equation number one was a + 2d is equal to 15. d we got as -1.
Now a + - -2 is = 15. So a is 15 + 2. So it's 17. Now we need to find the sum s 10 and the question it was find a 10. So this I'm waiting for you find out the value of a 10. So writing all the things a 10. So n easily you'll be knowing what is the value of n. A is 17 b is minus1.
Calculate it. So do it quickly. So this is for you. Solve it then we'll uh proceed further.
Explain once again this sum. Okay, I'll be explaining you.
Okay, repeating once again. So here the question was given as A3 A3 is given 15. Okay A3 is given 15 and S10 is given 125.
We need to find the value of D and A 10.
So from this first condition I prepared first equation. A3 is given 15. So a3 is equal to a + n -1 d. So replacing so a + n with uh replacing n with 3 as a3 is given. So replace n with 3 a + 3 - 1 d is equal to 15.
a + 3 - 1 is 2. a + 2d is 15. So this is first equation. Now sum of 10 terms is given 125. So formula of sum is n /x 2 within bracket 2 a + n -1 d is equal to 125.
So as sum of 10 terms is given n is 10 10 by 2 within bracket 2 a + n is 10. Now as sum of 10 terms. So n is 10. 10 - 1 d is equal to 125. Now two 5s are 10.
Here we can cancel out this. So first I canled 10 with 2 and with this five I canled 125. So here 125 25 are um sorry five 5 2's are 10 5 are 25. So with five I cancelled 125.
Now the equation is as follows. 2 a + 10 - 1 is 9. 2 a + 9 b is equal to 25. So we got two equations. First equation was a + 2d is equal to 15. Second is 2 a + 9 is equal to 25.
I don't know whether you know that linear equation in two variables or not.
So in linear equation uh linear equation in two variables we need to make any of the coefficient equal either we need to make the value of a equal or d equal. So here in the first equation a + 2d is there and the second equation 2 a + 9 d is there. If you want to make a equal whatever the value of a is there in the second equation that we will multiply to first and whatever there in the first equation that we will multiply to second equation. So here in the first equation only uh nothing is there so only one nothing means one. So here I multiplied nothing and then two is there. So the second equation I multiplied by two. So finally 2 a 2 2's are 4 d is equal to 15 2's are 30.
Second equation as it is 2 a + 9 d is equal to 25. Now changing sign so you have minus so 2 a got cancelled 4 - 9 is - 5. So it's - 5d is equal to 30 - 25 is 5. Now five and five got cancel. So d is minus1. We got the value of d.
So this value of d we need to put in any one of the equation to find out a. So that I placed in first equation. So first equation was like a + 2d a + 2d is equal to 15. So a + 2 * d s -1 is equal to 15. So a - 2 is equal to 15.
So a is 15 + 2 17. We got the value of a and d. Now here finally they gave the question find the value of a 10. So what is the formula of a 10? How you can calculate a 10? a is given b is given.
You need to find a 10.
Is it clear? Will you solve it by your own now?
Is it clear?
H. Now say answer quickly. What is the sum uh what is uh the value of a 10?
Find a 10.
so answer you got as eight. So let's check it once.
So I think maybe wrong.
So s a 10 is equal to a + n -1 d where n is 10. A is given 17 10 - 1 common difference is - 1 17 10 - 1 is 9 so it's + - 17 - 9 very good correct answer uh 8. Now one more question for you.
So here, A is given 2.
Common difference is 8.
Sn is given 90.
You need to find N and A N. So here find n and a n will you try or you want me to explain.
So here a is given d is given you need to find n. sn is given. So first use some formula to find out the value of n.
Then find out a n.
Okay, try it.
Explain John. Explain.
And this is going to be the last question.
I think for Sn is n / 2 2 a + n -1 d s 90 given nulate a is given 2 + n -1 common difference is given 8 questions factorization N / 2 2's are 4 + 8 N - 8 N / 2 8 N 4 4 - 8 is - 4 now it's given 90 2 so taking 2 common 4 - 2 2 common Cancellation sum simple to 90 is equal to n * 4 is 4 n² - 2 n. Now shifting all together the shift 4 n² - 2 n - 90 is = 0. Again two can be taken common two common so 2 n² - n - 45 is equal to 0 two zeros multiply 0 the final equation 2 n² - n - 45 is = 0. Now we are solving by middle term factorization 45 2's are 90k factors 10 9 are 90 10 - 9 is 1 bigger number sign of question middle factors 2 n² - 10 n + 9 n - 45 = 0 common n - 5 or say 9 commonly same bracket repeat n - 5 is equal to 0. So n - 5 is one of the factor 2 n + 9 is another factor. to n = 5 is opposite sign - 5 to - 5 opposite or n + 9 opposite 9 cannot be fraction value n can never be negative. So value of n is five. We got value of n question find n find a n find so a formula already a n is a + n -1 nulated as 5 to a 5.
So it's a a is 2 + 5 - 1 common difference is 8.
So 2 + 5 - 1 is 4 4 8s are 32.
2 + 32. So final answer is 34.
So a n is 34 and n is 5.
Okay? Is it clear now?
WhatsApp it will be easy for you.
Okay. Now I'm winding up the class. So once send out uh send me exercise from your textbook so that we can solve those questions.
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