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- 1. Arithmetic progressions Submitted to-mrs.archana savita
- 2. An Arithmatic Progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except first term. For example : 5, 10, 15, 20, 25….. In this each term is obtained by adding 5 to the preceding term except first term
- 3. The general form of an Arithmatic Progression is a , a +d , a + 2d , a + 3d ………………, a + (n-1)d The general form of an Arithmatic Progression is a , a +d , a + 2d , a + 3d ………………, a + (n-1)d Where ‘a’ is first term and ‘d’ is called common difference. Where ‘a’ is first term and ‘d’ is called common difference. nth Term of an A.P. is denoted by An An = a + (n-1)d nth Term of an A.P. is denoted by An An = a + (n-1)d
- 4. The first term = a1 =a +0 d = a + (1-1)d Let us consider an A.P. with first term ‘a’ and common difference ‘d’ ,then To find the nth Term of an A.P. The second term = a2 = a + d = a + (2-1)d The third term = a3 = a + 2d = a + (3- 1)dThe fourth term = a4 =a + 3d = a + (4-1)d ------------------------------------------- ------------------------------------------- The nth term = an = a + (n-1)d
- 5. To check that a given term is in A.P. or not. (i) 2, 6, 10, 14…. (i) Here first term a = 2, find differences in the next terms a2-a1 = 6 – 2 = 4 a3-a2 = 10 –6 = 4 a4-a3 = 14 – 10 = 4 Since the differences are common. Hence the given terms are in A.P.
- 6. Which of the following are in A.P. ? If they are in A.P. find their first term and common difference (i) 1, 3, 9, 27….. (ii) a, 2a, 3a, 4a… (iii) -6, -2, 2, 6…. (iv) 12 , 22 , 32 , 42 ……
- 7. Problem : Find the value of k for which the given series is in A.P. 4, k –1 , 12 Solution : Given A.P. is 4, k –1 , 12 If series is A.P. then the differences will be common d1 = d1 a2 – a1 = a3 – a2 k – 1 – 4 = 12 – (k – 1) k – 5 = 12 – k + 1 k + k = 12 + 1 + 5 2 k = 18 or k = 9
- 8. Problem 1. Find 10th term of A.P. 12, 18, 24, 30.. … Solution. Given A.P. is 12, 18, 24, 30.. First term is a = 12 Common difference is d = 18- 12 = 6 nth term is an = a + (n-1)d Put n = 10, a10 = 12 + (10-1)6 = 12 + 9 x 6 = 12 + 54 a10 = 66
- 9. Problem 2. Find number of terms of A.P. 100, 105, 110, 115,,………………500 Solution. Given A.P. is 100, 105, 110, 115,………………500 First term is a = 100 , an = 500 Common difference is d = 105 -100 = 5 nth term is an = a + (n-1)d 500 = 100 + (n-1)5 500 - 100 = 5(n – 1) 400 = 5(n – 1) 5(n – 1) = 400
- 10. Problem 3. Find the number of all three digit numbers divisible by 8 Solution. Smallest three digit number divisible by 8 is 104 & Largest three digit number divisible by 8 is 992 Therefore numbers obtained are 104, 112,120, 128, …………………992 Which are in A.P. Now solve using A.P. and find how many are they ?
- 11. 5(n – 1) = 400 n – 1 = 400/5 n - 1 = 80 n = 80 + 1 n = 81 Hence the no. of terms are 81.
- 12. Let us solve some problems based on Arithmatic Progression 1. Find 50th term of A.P. 5, 10, 15, 20… 2. Find first three terms of an A.P. whose 3rd term is 10 and 6th term is 19. 3. Find number of terms of an A.P. 3, 6, 9, ………………..99.
- 13. Sum of n terms of an Arithmetic Progression Its formula is Sn = ½ n [ 2a + (n - 1)d ] It can also be written as Sn = ½ n [ a + an ]
- 14. You may need to be able to prove this formula. It is derived as follows: The sum to n terms is given by: Sn = a + (a + d) + (a + 2d) + … + (a + (n – 1)d) (1) If we write this out backwards, we get: Sn = (a + (n – 1)d) + (a + (n – 2)d) + … + a (2) The sum to n terms of an arithmetic progression Sn = ½ n [ 2a + (n - 1)d ]
- 15. Now let’s add (1) and (2): 2Sn = [2a + (n – 1)d] + [2a + (n – 1)d] + … ……….. + [2a + (n – 1)d] So Sn = ½ n [2a + (n – 1)d]
- 16. Problem 1. Find the sum of 30 terms of given A.P. 12 + 20 + 28 + 36……… Solution : Given A.P. is 12 , 20, 28 , 36 Its first term is a = 12 Common difference is d = 20 – 12 = 8 The sum to n terms of an arithmetic progression Sn = ½ n [ 2a + (n - 1)d ] = ½ x 30 [ 2x 12 + (30-1)x 8] = 15 [ 24 + 29 x8] = 15[24 + 232] = 15 x 246 = 3690
- 17. Problem 2. Find the sum given A.P. 2 + 4 + 6 + 8 + ……………… + 200 Solution : Given A.P. is 2 , 4, 6 , 8 …………….200 Its first term is a = 2 Common difference is d = 4 – 2 = 2 nth term is an = a + (n-1)d 200 = 2 + (n-1)2 200 - 2 = 2(n – 1) 198 = 2(n – 1) 2(n – 1) = 198
- 18. The sum to n terms of an arithmetic progression Sn = ½ n [ 2a + (n - 1)d ] S100 = ½ x 100 [ 2x 2 + (100-1)x 2] = 50 [ 4 + 198] = 50[202] = 10100 2(n – 1) = 198 n – 1 = 198/2 n – 1 = 99 n = 99 + 1 n = 100
- 19. Let us solve some problems based on sum of terms of Arithmatic Progression 1. Find sum of 40 term of A.P. 5, 10, 15, 20… 3. Find the sum of series 10 + 20 + 30 +… …………+ 1000 2. Find sum of 100 term of A.P. -70, -67, -64… 4. Find the sum of series -5 + (-15)+(-25+… …………+ (-525)
- 20. ARITHMETIC SEQUENCES These are sequences where the difference between successive terms of a sequence is always the same number. This number is called the common difference.
- 21. 3, 7, 11, 15, 19 … Notice in this sequence that if we find the difference between any term and the term before it we always get 4. 4 is then called the common difference and is denoted with the letter d. d = 4 To get to the next term in the sequence we would add 4 so a recursive formula for this sequence is: 41 += −nn aa The first term in the sequence would be a1 which is sometimes just written as a. a = 3
- 22. 3, 7, 11, 15, 19 … +4 +4 +4 +4 Each time you want another term in the sequence you’d add d. This would mean the second term was the first term plus d. The third term is the first term plus d plus d (added twice). The fourth term is the first term plus d plus d plus d (added three times). So you can see to get the nth term we’d take the first term and add d (n - 1) times. d = 4 ( )dnaan 1−+= Try this to get the 5th term. a = 3 ( ) 1916341535 =+=−+=a
- 23. Let’s look at a formula for an arithmetic sequence and see what it tells us. { }14 −n Subbing in the set of positive integers we get: 3, 7, 11, 15, 19, … What is the common difference? d = 4 you can see what the common difference will be in the formula We can think of this as a “compensating term”. Without it the sequence would start at 4 but this gets it started where we want it. 4n would generate the multiples of 4. With the - 1 on the end, everything is back one. What would you do if you wanted the sequence 2, 6, 10, 14, 18, . . .? { }24 −n
- 24. Find the nth term of the arithmetic sequence when a = 6 and d = -2 If we use -2n we will generate a sequence whose common difference is -2, but this sequence starts at -2 (put 1 in for n to get first term to see this). We want ours to start at 6. We then need the “compensating term”. If we are at -2 but want 6, we’d need to add 8. { }82 +− n Check it out by putting in the first few positive integers and verifying that it generates our sequence. 6, 4, 2, 0, -2, . . . Sure enough---it starts at 6 and has a common difference of -2
- 25. Let’s try something a little trickier. What if we just know a couple of terms and they aren’t consecutive? The fourth term is 3 and the 20th term is 35. Find the first term and both a term generating formula and a recursive formula for this sequence. How many differences would you add to get from the 4th term to the 20th term? daa 16420 +=35 3 Solve this for d d = 2 The fourth term is the first term plus 3 common differences. daa 314 +=3 (2) 31 −=a We have all the info we need to express these sequences. We’ll do it on next slide. 35,3 204 == aa
- 26. The fourth term is 3 and the 20th term is 35. Find the first term and both a term generating formula and a recursive formula for this sequence. 35,3 204 == aa d = 2 31 −=a { }52 −n makes the common difference 2 makes the first term - 3 instead of 2 21 += −nn aaThe recursive formula would be: Let’s check it out. If we find n = 4 we should get the 4th term and n = 20 should generate the 20th term. ( ) 35424 =−=a ( ) 35520220 =−=a
- 27. Often in applications we will want the sum of a certain number of terms in an arithmetic sequence. The story is told of a grade school teacher In the 1700's that wanted to keep her class busy while she graded papers so she asked them to add up all of the numbers from 1 to 100. These numbers are an arithmetic sequence with common difference 1. Carl Friedrich Gauss was in the class and had the answer in a minute or two (remember no calculators in those days). This is what he did: 1 + 2 + 3 + 4 + 5 + . . . + 96 + 97 + 98 + 99 + 100 sum is 101 sum is 101 With 100 numbers there are 50 pairs that add up to 101. 50(101) = 5050
- 28. This will always work with an arithmetic sequence. The formula for the sum of n terms is: ( )nn aa n S += 1 2 n is the number of terms so n/2 would be the number of pairs first term last term Let’s find the sum of 1 + 3 +5 + . . . + 59But how many terms are there? We can write a formula for the sequence and then figure out what term number 59 is.
- 29. ( )nn aa n S += 1 2 first term last term Let’s find the sum of 1 + 3 +5 + . . . + 59 { }12 −n The common difference is 2 and the first term is one so: Set this equal to 59 to find n. Remember n is the term number. 2n - 1 = 59 n = 30So there are 30 terms to sum up. ( ) 900591 2 30 30 =+=S

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