Wikipedia
Arithmetic progression - In mathematics, an arithmetic progression or arithmetic sequence is a sequence of number s such that the difference of any two successive members of the sequence is a constant. For instance, the sequence 3, 5, 7, 9, 11, 13, … is an arithmetic progression with common difference 2...
Arithmetic progression - In mathematics, an arithmetic progression (A.P.) or arithmetic sequence is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. For instance, the sequence 3, 5, 7, 9, 11, 13, … is an arithmetic progression with common difference 2. If the initial term of an arithmetic progression is and the common difference of successive members is d, then the n th term of the sequence is given by: and in general A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes just called an arithmetic progression. The sum of the members of a finite arithmetic progression is called an arithmetic series. Express the arithmetic series in two different ways: Add both sides of the two equations. All terms involving d cancel, and so we're left with: Rearranging and remembering that , we get: This formula has long been known, but Carl Friedrich Gau....
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Arithmetic Progression
Quantities are said to be in Arithmetic progression when they increase or decrease by a common differenc..
Arithmetic Progression- Test Questions
Question 11 - Question: Find three numbers in AP whose sum is 24 and whose product is 440. Answer: Let a-d, a, a+d be three numbers in AP Then (a - d) + (a) + (a + d) = 24 3a = 24 a = 8 (a - d)(a)(a + d) = 440 (8 - d)(8)(8 + d) = 4..
Question 11 - Question: Find three numbers in AP whose sum is 24 and whose product is 440. Answer: Let a-d, a, a+d be three numbers in AP Then (a - d) + (a) + (a + d) = 24 3a = 24 a = 8 (a - d)(a)(a + d) = 440 (8 - d)(8)(8 + d) = 4..Problems on Arithmetic Progression - Test Questions
Question 21 - Question: (pq) t h term and the sum of pq terms. Answer: Let a = first term, d = common difference Subtract (ii) from (..
Question 21 - Question: (pq) t h term and the sum of pq terms. Answer: Let a = first term, d = common difference Subtract (ii) from (..Arithmetic Progression (or simply A.P.)
Arithmetic Progression (or simply A.P.) - Quantities are said to be in Arithmetic progression when they increase or decrease by a common differen..
How do you solve this arithmetic progression Q?CBSE Maths 10th Arithmetic Progression
Is Sierpinski's Triangle based on Arithmetic Progression?more like a Blair from GG or Brooke from OTH, more girly, outgoing etc... =P Note for siriusblackbabe's contest: triangle category. Please add/subscribe to my back-up account: www.youtube.com and my little sister's account: www.youtube.com Thaaaank yooou!! Loveee yoou all!! xD 2nd Place in MariaPurt's 'Strong Female Character' multifandom CONTEST www.youtube.com 2nd Place in the Triangles Category in siriusblackbabe's BIG MULTI-FANDOM CONTEST!!! (out of almost 200 videos) www.youtube.com ...
Question : An arithmetic series consists of 35 natural numbers, that sum up to a total of 1890. Prove that 25 cannot be a number in that series.
Answer : I'm afraid that while everybody seems to have some good ideas, I can answer this with an answer better and more simple than that of Stephen B or anybody else. It's quite easy. There are an odd number of terms in the progression. That means the middle term is the average: 1890 / 35 = 54 For 25 to be included in the series, the gap (usually called d) in the progression must divide the difference: 54 - 25 = 29 For that to be true, the gap would either be 1 or 29. It cannot be 29, because that causes us to have negative numbers in our progression. It cannot be 1, because with only 35 terms, there are only 17 above and 17 below the average - it cannot extend low enough to reach the 25. This solution is the best and most elegant. In fact, it's the only complete proof given (the rest seem to end with ... or 'etc' or something like that, leaving you to write out a tedious, unelegant solution).
Answer : I'm afraid that while everybody seems to have some good ideas, I can answer this with an answer better and more simple than that of Stephen B or anybody else. It's quite easy. There are an odd number of terms in the progression. That means the middle term is the average: 1890 / 35 = 54 For 25 to be included in the series, the gap (usually called d) in the progression must divide the difference: 54 - 25 = 29 For that to be true, the gap would either be 1 or 29. It cannot be 29, because that causes us to have negative numbers in our progression. It cannot be 1, because with only 35 terms, there are only 17 above and 17 below the average - it cannot extend low enough to reach the 25. This solution is the best and most elegant. In fact, it's the only complete proof given (the rest seem to end with ... or 'etc' or something like that, leaving you to write out a tedious, unelegant solution).
Question : If the sum of 7 terms is 49 and the sum of 17 terms is 289,find the sum of n terms.
solve in detail
Answer : Hi, Sn = n/2(2a + d(n - 1)) For 7 terms this becomes: 49 = 7/2(2a + d(7 - 1)) 49 = 7/2(2a + 6d) 98 = 14a + 42d Sn = n/2(2a + d(n - 1)) For 17 terms this becomes: 289 = 17/2(2a + d(17 - 1)) 289 = 17/2(2a + 16d) 578 = 34a + 272d This gives 2 equations 578 = 34a + 272d 98 = 14a + 42d Multiply the top equation by 7 and the bottom equation by -17 and add them. 7(578 = 34a + 272d) -17(98 = 14a + 42d) 4046 = 238a + 1904d -1666 = -238a - 714d -------------------------------- 2380 = 1190d 2 = d If d = 2 and 98 = 14a + 42d, then 98 = 14a + 42(2) 98 = 14a + 84 14 = 14a 1 = a The Sum of n terms would be: Sn = n/2(2*1 + 2(n - 1)) this simplifies to: Sn = n/2(2 + 2n - 2) Sn = n/2(2n) Sn = n <== ANSWER I hope that helps!! :-)
Answer : Hi, Sn = n/2(2a + d(n - 1)) For 7 terms this becomes: 49 = 7/2(2a + d(7 - 1)) 49 = 7/2(2a + 6d) 98 = 14a + 42d Sn = n/2(2a + d(n - 1)) For 17 terms this becomes: 289 = 17/2(2a + d(17 - 1)) 289 = 17/2(2a + 16d) 578 = 34a + 272d This gives 2 equations 578 = 34a + 272d 98 = 14a + 42d Multiply the top equation by 7 and the bottom equation by -17 and add them. 7(578 = 34a + 272d) -17(98 = 14a + 42d) 4046 = 238a + 1904d -1666 = -238a - 714d -------------------------------- 2380 = 1190d 2 = d If d = 2 and 98 = 14a + 42d, then 98 = 14a + 42(2) 98 = 14a + 84 14 = 14a 1 = a The Sum of n terms would be: Sn = n/2(2*1 + 2(n - 1)) this simplifies to: Sn = n/2(2 + 2n - 2) Sn = n/2(2n) Sn = n <== ANSWER I hope that helps!! :-)