Arithmetic series. How do you find the values of the first number and the common difference?
Q. In an arithmetic series, how do you find the values of the first number and the common difference from the sum of two terms in the series?
Asked by creativecontrol - Tue Apr 28 03:46:40 2009 - - 3 Answers - 0 Comments
A. If you know two terms and what their nth value is, i.e. - the 5th term in a series is 6, and the 7th term in the series is 9. Then the nth term of an arithmetic progression is a + (n-1)d where a is the first term and d is the common difference. So 5th term --> a + 4d = 6 So 7th term --> a + 6d = 9 Then, you would have one possibility --> Subtract the 1st from the 2nd so: (2) - (1) 2d = 3, d = 3/2 Would be the common difference for the series in question and you would plug is back into say, a + 4(3/2) = 6 a + 6 = 6 a = 0, Mind you, this is just an example. But we have formulas for the sum of an arithmetic series up to a certain term - i.e. the sum of the first 10 terms, but you can't find it by just using the sum of two terms,… [cont.]
Answered by Shan Bokhari - Tue Apr 28 05:01:13 2009
Q. In an arithmetic series, how do you find the values of the first number and the common difference from the sum of two terms in the series?
Asked by creativecontrol - Tue Apr 28 03:46:40 2009 - - 3 Answers - 0 Comments
A. If you know two terms and what their nth value is, i.e. - the 5th term in a series is 6, and the 7th term in the series is 9. Then the nth term of an arithmetic progression is a + (n-1)d where a is the first term and d is the common difference. So 5th term --> a + 4d = 6 So 7th term --> a + 6d = 9 Then, you would have one possibility --> Subtract the 1st from the 2nd so: (2) - (1) 2d = 3, d = 3/2 Would be the common difference for the series in question and you would plug is back into say, a + 4(3/2) = 6 a + 6 = 6 a = 0, Mind you, this is just an example. But we have formulas for the sum of an arithmetic series up to a certain term - i.e. the sum of the first 10 terms, but you can't find it by just using the sum of two terms,… [cont.]
Answered by Shan Bokhari - Tue Apr 28 05:01:13 2009
Can You Help Me With Arithmetic Series Please ?
Q. The 2nd and 5th terms of an arithmetic series are 26 and 41 respectively. The common difference is 5. The 12th term of the series is 76. Another arithmetic series has 1st term -12 and common difference 7. Given that the sums of the first n terms of these two series are equal, - Find the value of n.
Asked by Vicky S - Tue Nov 18 14:03:17 2008 - - 2 Answers - 0 Comments
A. Arithmetic series formula: a(n) = a(1) +(n-1)d a(2) 26 so: 26 = a(1) +(2-1)*5 26 = a(1) +5 a(1)= 21 we also know 41 = a(1) +(5-1)*5 41 = a(1) +20 a(1) = 21 So there is no contradiction with two bits of information regarding the series. The other series: a(n) = -12 +(n-1)*7 If the sum of the first n terms is equal we will use the sum formula S =n/2*(a(1) +a(n)) or S =n/2*(a(1) +a(1)+(n-1)*d) n/2(42+(n-1)*5)=n/2(-12+( n-1)*7) 42+5(n-1) = -12+7(n-1) 37 +5n =-19+ 7n 2n = 56 n = 28
Answered by Peter m - Tue Nov 18 14:16:03 2008
Q. The 2nd and 5th terms of an arithmetic series are 26 and 41 respectively. The common difference is 5. The 12th term of the series is 76. Another arithmetic series has 1st term -12 and common difference 7. Given that the sums of the first n terms of these two series are equal, - Find the value of n.
Asked by Vicky S - Tue Nov 18 14:03:17 2008 - - 2 Answers - 0 Comments
A. Arithmetic series formula: a(n) = a(1) +(n-1)d a(2) 26 so: 26 = a(1) +(2-1)*5 26 = a(1) +5 a(1)= 21 we also know 41 = a(1) +(5-1)*5 41 = a(1) +20 a(1) = 21 So there is no contradiction with two bits of information regarding the series. The other series: a(n) = -12 +(n-1)*7 If the sum of the first n terms is equal we will use the sum formula S =n/2*(a(1) +a(n)) or S =n/2*(a(1) +a(1)+(n-1)*d) n/2(42+(n-1)*5)=n/2(-12+( n-1)*7) 42+5(n-1) = -12+7(n-1) 37 +5n =-19+ 7n 2n = 56 n = 28
Answered by Peter m - Tue Nov 18 14:16:03 2008
How do I solve and find the n term for partial sums for an arithmetic series?
Q. Arithmetic Series : Sn= 3219 t1=15 d=4 Find n. Arithmetic Series : Sn=25477.9 t1=1.7 t2=5.3 Find n.
Asked by ally c - Thu Apr 3 21:08:33 2008 - - 1 Answers - 0 Comments
A. Knowing that Sn=n[2 t1 + (n-1) d] /2 you get 3219=n[30 + (n-1) 4] /2, solve for n and you will find that n=37 For your second series you can apply the same formula for Sn, but need to calculate d at first, which can be easily done, as d=t2-t1=5.3-1.7=3.6 Then Sn=n[2 t1 + (n-1) d] /2 becomes 25477.9 = n[3.4 + (n-1) 3.6] /2 This is true for n=119
Answered by Stefan E - Fri Apr 4 14:25:12 2008
Q. Arithmetic Series : Sn= 3219 t1=15 d=4 Find n. Arithmetic Series : Sn=25477.9 t1=1.7 t2=5.3 Find n.
Asked by ally c - Thu Apr 3 21:08:33 2008 - - 1 Answers - 0 Comments
A. Knowing that Sn=n[2 t1 + (n-1) d] /2 you get 3219=n[30 + (n-1) 4] /2, solve for n and you will find that n=37 For your second series you can apply the same formula for Sn, but need to calculate d at first, which can be easily done, as d=t2-t1=5.3-1.7=3.6 Then Sn=n[2 t1 + (n-1) d] /2 becomes 25477.9 = n[3.4 + (n-1) 3.6] /2 This is true for n=119
Answered by Stefan E - Fri Apr 4 14:25:12 2008
Can someone help me with Arithmetic series and sequences?
Q. Hi, I'm having some trouble with these. These are just to refresh my memory but I lost my notes and am trying to remember! 1) Find the sum of the n terms of the arithmetic sequence a1 = 7, a12 = 29 and n = 12. 2) Find the sum of the first 50 terms 1, 8, 15... using the sum of an arithmetic series formula thanks!
Asked by Flying High - Thu Aug 30 20:30:13 2007 - - 1 Answers - 0 Comments
A. a12 = a1 + 11r (where r is the amount added to each term to get the next). We know 11 of them are added between the first and twelfth. 29 = 11r + 7 r = 2. sum(a1...a12) = sum(a1, a1+r, a1+2r,...,a1+11r) = 12a1 + r sum(1,2,...,11) = 12a1 + r(11*12/2) = 12a1 + 66r = 12*7 + 66*2 = 216 to check, sum = 6(a1+a12) = 6*36 = 216 a1 = 1, r = 7, n = 50 sum = 50*1 + 7 * (49*50/2) = 8625
Answered by holdm - Thu Aug 30 20:42:53 2007
Q. Hi, I'm having some trouble with these. These are just to refresh my memory but I lost my notes and am trying to remember! 1) Find the sum of the n terms of the arithmetic sequence a1 = 7, a12 = 29 and n = 12. 2) Find the sum of the first 50 terms 1, 8, 15... using the sum of an arithmetic series formula thanks!
Asked by Flying High - Thu Aug 30 20:30:13 2007 - - 1 Answers - 0 Comments
A. a12 = a1 + 11r (where r is the amount added to each term to get the next). We know 11 of them are added between the first and twelfth. 29 = 11r + 7 r = 2. sum(a1...a12) = sum(a1, a1+r, a1+2r,...,a1+11r) = 12a1 + r sum(1,2,...,11) = 12a1 + r(11*12/2) = 12a1 + 66r = 12*7 + 66*2 = 216 to check, sum = 6(a1+a12) = 6*36 = 216 a1 = 1, r = 7, n = 50 sum = 50*1 + 7 * (49*50/2) = 8625
Answered by holdm - Thu Aug 30 20:42:53 2007
Write the series using sigma notation, an arithmetic series with 11 terms with 1st term 8 and last term 508?
Q. Write the series using sigma notation, an arithmetic series with 11 terms with 1st term 8 and last term 508. How do I do that???
Asked by pockyeaters - Mon Apr 14 19:23:12 2008 - - 1 Answers - 0 Comments
A. 1st term = 8. 11th term = 508. Therefore, common addend = (508-8)/(11 - 1) = 500/10 = 50. In sigma notation, you have... n = 0 to 10 (8 + 50n). -John
Answered by John - Mon Apr 14 19:32:50 2008
Q. Write the series using sigma notation, an arithmetic series with 11 terms with 1st term 8 and last term 508. How do I do that???
Asked by pockyeaters - Mon Apr 14 19:23:12 2008 - - 1 Answers - 0 Comments
A. 1st term = 8. 11th term = 508. Therefore, common addend = (508-8)/(11 - 1) = 500/10 = 50. In sigma notation, you have...
Answered by John - Mon Apr 14 19:32:50 2008
Arithmetic series question - an arithmetic series has first term a and common difference d.?
Q. a) Prove that the sum of the first n terms of the series is equal to na+1/2n(n-1)d. The third term of an arithmetic series is -7. The eight terms from the fourth to eleventh term inclusive have a sum of 70. b) Calculate the first term and the common difference of this series
Asked by todster - Sun Oct 18 13:18:37 2009 - - 2 Answers - 0 Comments
A. let S = a + (a + d) + (a + 2d)...[ a + (n - 2) d ] + [a + (n-1)d ] write in the reverse order S = [a + (n-1)d ] + [ a + (n - 2) d ]...(a+2d) + (a + d) + a add above two eqns 2S = [2a + (n-1)d ] + [2a + (n-1)d ] ...[ 2a + (n - 1) d ] 2S = n [ 2a + (n - 1) d] divide by 2 S = na + (1/2)(n - 1)d b) a_3 = -7 a_4 + a_5...a_11 = 70 a_3 + a_4 + a_5...a_11 = 63 since S_n = n/2 [ 2a + (n - 1)d ] sum of 9 terms from third to 11 is 9/2 [ 2*a_3 + (9 - 1)d ] = 63 9a_3 + 36d = 63 substitute a_3 = -7 -63 + 36d = 63 36d = 126 d = 126/36 = 7/2 since a_3 = a_1 + 2d a_1 + 7 = -7 a_1 = -14 first term = -14 and common difference = 7/2
Answered by mohanrao d - Sun Oct 18 13:37:28 2009
Q. a) Prove that the sum of the first n terms of the series is equal to na+1/2n(n-1)d. The third term of an arithmetic series is -7. The eight terms from the fourth to eleventh term inclusive have a sum of 70. b) Calculate the first term and the common difference of this series
Asked by todster - Sun Oct 18 13:18:37 2009 - - 2 Answers - 0 Comments
A. let S = a + (a + d) + (a + 2d)...[ a + (n - 2) d ] + [a + (n-1)d ] write in the reverse order S = [a + (n-1)d ] + [ a + (n - 2) d ]...(a+2d) + (a + d) + a add above two eqns 2S = [2a + (n-1)d ] + [2a + (n-1)d ] ...[ 2a + (n - 1) d ] 2S = n [ 2a + (n - 1) d] divide by 2 S = na + (1/2)(n - 1)d b) a_3 = -7 a_4 + a_5...a_11 = 70 a_3 + a_4 + a_5...a_11 = 63 since S_n = n/2 [ 2a + (n - 1)d ] sum of 9 terms from third to 11 is 9/2 [ 2*a_3 + (9 - 1)d ] = 63 9a_3 + 36d = 63 substitute a_3 = -7 -63 + 36d = 63 36d = 126 d = 126/36 = 7/2 since a_3 = a_1 + 2d a_1 + 7 = -7 a_1 = -14 first term = -14 and common difference = 7/2
Answered by mohanrao d - Sun Oct 18 13:37:28 2009
What is the sum of the following arithmetic series: 2+3+4+...+108?
Q. The word question is: For each of her birthdays, Ann blew out the number of candles equal to her age, plus one for good luck. If she lived to 107, what is the total number of candles she would have blown out over her lifetime?
Asked by Janice - Sun Oct 25 20:12:00 2009 - - 4 Answers - 0 Comments
Q. The word question is: For each of her birthdays, Ann blew out the number of candles equal to her age, plus one for good luck. If she lived to 107, what is the total number of candles she would have blown out over her lifetime?
Asked by Janice - Sun Oct 25 20:12:00 2009 - - 4 Answers - 0 Comments
Need help finding terms of arithmetic series?
Q. Find the 10th term of the series, s10, of an arithmetic series whose 3rd term, s3, is 7 and whose 43rd term, s43, is 17. I tried using, an = am r^(n-m) , but I kept getting 8.75. Here are the possible answers... A. 42.25 B. 59.32 C. 62.5 D. 76.25 If you wouldn't mind showing the process...thanks!
Asked by WK - Sun Jun 7 16:45:03 2009 - - 1 Answers - 0 Comments
A. You mixed up arithmetic and geometric series. :) More precisely, you mixed up ^ and *. :) That said, the answer needs to be between 7 and 17 in any case, which none of those possibilities are, so obviously there's some other typo as well. I suggest you reread the problem carefully and try again. :D
Answered by Curt Monash - Sun Jun 7 18:46:26 2009
Q. Find the 10th term of the series, s10, of an arithmetic series whose 3rd term, s3, is 7 and whose 43rd term, s43, is 17. I tried using, an = am r^(n-m) , but I kept getting 8.75. Here are the possible answers... A. 42.25 B. 59.32 C. 62.5 D. 76.25 If you wouldn't mind showing the process...thanks!
Asked by WK - Sun Jun 7 16:45:03 2009 - - 1 Answers - 0 Comments
A. You mixed up arithmetic and geometric series. :) More precisely, you mixed up ^ and *. :) That said, the answer needs to be between 7 and 17 in any case, which none of those possibilities are, so obviously there's some other typo as well. I suggest you reread the problem carefully and try again. :D
Answered by Curt Monash - Sun Jun 7 18:46:26 2009
An Arithmetic series has a first term of 3 and a fifth term of 39. a geometric series has a common ratio of 3.
Q. The sum of the first two terms of the geometric series is the same as the second term of the arithmetic series. what is the first term of the geometric series?
Asked by 12234 - Sat Dec 1 05:29:27 2007 - - 3 Answers - 0 Comments
A. Arithmetic series: t_1 = 3; t_5 = 39 = 3 + d(5 - 1) or d = 9 So, t_2 = 3 + 9 = 12 Geometric series: t_1 + t_2 = 12 Also, t_2 = 3*(t_1) So, t_1 + 3(t_1) = 12 or t_1 = 3
Answered by psbhowmick - Sat Dec 1 05:41:26 2007
Q. The sum of the first two terms of the geometric series is the same as the second term of the arithmetic series. what is the first term of the geometric series?
Asked by 12234 - Sat Dec 1 05:29:27 2007 - - 3 Answers - 0 Comments
A. Arithmetic series: t_1 = 3; t_5 = 39 = 3 + d(5 - 1) or d = 9 So, t_2 = 3 + 9 = 12 Geometric series: t_1 + t_2 = 12 Also, t_2 = 3*(t_1) So, t_1 + 3(t_1) = 12 or t_1 = 3
Answered by psbhowmick - Sat Dec 1 05:41:26 2007
How are arithmetic series & sequences used in everday life?
Q. Such as the zeta regularized equations or arithmetic progression .. how would someone use them in their career and what purpose would it serve.
Asked by k.r.e.a.m - Wed Jun 13 09:46:09 2007 - - 1 Answers - 0 Comments
A. Arithemtic sequences can easily be seen for things that have a one-time cost, and then a secondary periodic cost such as: 1) Telephone bill -- basic monthly cost + cost for minutes used 2) Taking a group a distance to an event -- gas to get there + admission per person 3) Going to a club -- cover charge + cost per drink You get the idea. Now, as a consumer, it's not always neccessary to figure out how much things will be ahead of time (beyond a good estimate of the final total cost), but it can help you to get the best deal if the savings will be worth your time to do the math. However, if you own the store, it can help you figure out how to increase your bottom line because you can alter your pricing to give you and advantage over… [cont.]
Answered by math guy - Wed Jun 13 18:17:59 2007
Q. Such as the zeta regularized equations or arithmetic progression .. how would someone use them in their career and what purpose would it serve.
Asked by k.r.e.a.m - Wed Jun 13 09:46:09 2007 - - 1 Answers - 0 Comments
A. Arithemtic sequences can easily be seen for things that have a one-time cost, and then a secondary periodic cost such as: 1) Telephone bill -- basic monthly cost + cost for minutes used 2) Taking a group a distance to an event -- gas to get there + admission per person 3) Going to a club -- cover charge + cost per drink You get the idea. Now, as a consumer, it's not always neccessary to figure out how much things will be ahead of time (beyond a good estimate of the final total cost), but it can help you to get the best deal if the savings will be worth your time to do the math. However, if you own the store, it can help you figure out how to increase your bottom line because you can alter your pricing to give you and advantage over… [cont.]
Answered by math guy - Wed Jun 13 18:17:59 2007
Help finding the common difference in an arithmetic series?
Q. The 12th term of A is 42 and the sum of the first 17 terms is 51. I know that 42=A1+11d, and 51=17((A1+An)/n), I'm just not sure how to solve for the common difference. Any guidance would be greatly appreciated.
Asked by Jack Mania - Thu Apr 16 19:03:11 2009 - - 2 Answers - 0 Comments
A. 51=17((A1+An)/n), this equation is wrong, it should be 51=17(A1+An)/2 and A(17) = A1 + 16d ---> 2A1 + 16d = 6 and A1+11d = 42 solve 2 equations --->d = 13
Answered by kim t - Thu Apr 16 19:16:40 2009
Q. The 12th term of A is 42 and the sum of the first 17 terms is 51. I know that 42=A1+11d, and 51=17((A1+An)/n), I'm just not sure how to solve for the common difference. Any guidance would be greatly appreciated.
Asked by Jack Mania - Thu Apr 16 19:03:11 2009 - - 2 Answers - 0 Comments
A. 51=17((A1+An)/n), this equation is wrong, it should be 51=17(A1+An)/2 and A(17) = A1 + 16d ---> 2A1 + 16d = 6 and A1+11d = 42 solve 2 equations --->d = 13
Answered by kim t - Thu Apr 16 19:16:40 2009
I need help for this math 10 question about the sum of an arithmetic series?
Q. Yes i left my homwork to the very last minute. Here is the question: Raji's annual salary is in a range from $25 325 in the 1st year to $34 445 in the 7th year a) The salary range is an arithmetic sequence with seven terms. Determine the raise Raji can expect each year. The answer is $1520. Someone please help me, i don't get it D:!
Asked by Jasmine - Sun Sep 27 23:56:29 2009 - - 2 Answers - 0 Comments
A. you take the difference of the 2 salaries, in this case it is $9120, and divide it by 6 years, the number of years left till his seventh year. and presto you have $1520
Answered by 419spreadneck - Mon Sep 28 00:08:21 2009
Q. Yes i left my homwork to the very last minute. Here is the question: Raji's annual salary is in a range from $25 325 in the 1st year to $34 445 in the 7th year a) The salary range is an arithmetic sequence with seven terms. Determine the raise Raji can expect each year. The answer is $1520. Someone please help me, i don't get it D:!
Asked by Jasmine - Sun Sep 27 23:56:29 2009 - - 2 Answers - 0 Comments
A. you take the difference of the 2 salaries, in this case it is $9120, and divide it by 6 years, the number of years left till his seventh year. and presto you have $1520
Answered by 419spreadneck - Mon Sep 28 00:08:21 2009
find the sum of the arithmetic series given the first and last terms?
Q. this question tells me to find the sum of the arithmetic series given the first and last terms? the first term which is known as t1 is 6 and the last term which is known as t9 is 24 please help i have no idea how i should do this question
Asked by mshoker - Fri Dec 12 10:42:21 2008 - - 1 Answers - 0 Comments
A. sum=n/2(a+l) where n=numbr of terms a is the 1st term and l is the last term
Answered by apoorva - Fri Dec 12 10:52:28 2008
Q. this question tells me to find the sum of the arithmetic series given the first and last terms? the first term which is known as t1 is 6 and the last term which is known as t9 is 24 please help i have no idea how i should do this question
Asked by mshoker - Fri Dec 12 10:42:21 2008 - - 1 Answers - 0 Comments
A. sum=n/2(a+l) where n=numbr of terms a is the 1st term and l is the last term
Answered by apoorva - Fri Dec 12 10:52:28 2008
find the sum of the first n terms of the arithmetic series?
Q. an example is: 3+6+9+12+15+...: n=10 if you can answer this in simple terms please do. if not post a link to a website that explains it in simple terms. i have 7 of these problems to do, and i dont know how. no, i dont get it. you showed the steps but you didnt explain it at all. thanks for the jibberish.
Asked by oh no its me! - Wed Mar 25 00:08:46 2009 - - 1 Answers - 0 Comments
A. the formula is Sn=(d/2)n^2+ (U1- d/2)n your common difference (d) is 3 and your n is 10 so, S10=(3/2)10^2+(3-3/2)10 the sum of 10 terms is (1.5*100)+(1.5*10) 150+15=165 hope you get it.
Answered by wampleralex - Wed Mar 25 00:24:44 2009
Q. an example is: 3+6+9+12+15+...: n=10 if you can answer this in simple terms please do. if not post a link to a website that explains it in simple terms. i have 7 of these problems to do, and i dont know how. no, i dont get it. you showed the steps but you didnt explain it at all. thanks for the jibberish.
Asked by oh no its me! - Wed Mar 25 00:08:46 2009 - - 1 Answers - 0 Comments
A. the formula is Sn=(d/2)n^2+ (U1- d/2)n your common difference (d) is 3 and your n is 10 so, S10=(3/2)10^2+(3-3/2)10 the sum of 10 terms is (1.5*100)+(1.5*10) 150+15=165 hope you get it.
Answered by wampleralex - Wed Mar 25 00:24:44 2009
The sum of the first 6 term of an arithmetic series is 297 and the sum of the first 8 terms is 500?
Q. The sum of the first 6 term of an arithmetic series is 297 and the sum of the first 8 terms is 500. determine the sum of the first 3 terms, plz i need help. show me how to do it. thanks==
Asked by Mingsi - Fri Sep 18 00:29:15 2009 - - 3 Answers - 0 Comments
A. Sn= n/2[ 2a+(n-1)d] s6= 6/2[ 2a+ 5d]= 297 2a+5d= 99 A S8= 8/2[ 2a+7d]=500 2a+7d= 125 B Subtract A from B 2d= 125-99= 26 d= 13 2a+5d=99 2a= 99- 5d =99-5x 13 = 99-65 =34 S3= 3/2[ 2a+2d] = 3/2[ 34+ 26] =3/2 x 60 = 90 ie 17+30+ 43=90
Answered by DOVE - Fri Sep 18 00:44:42 2009
Q. The sum of the first 6 term of an arithmetic series is 297 and the sum of the first 8 terms is 500. determine the sum of the first 3 terms, plz i need help. show me how to do it. thanks==
Asked by Mingsi - Fri Sep 18 00:29:15 2009 - - 3 Answers - 0 Comments
A. Sn= n/2[ 2a+(n-1)d] s6= 6/2[ 2a+ 5d]= 297 2a+5d= 99 A S8= 8/2[ 2a+7d]=500 2a+7d= 125 B Subtract A from B 2d= 125-99= 26 d= 13 2a+5d=99 2a= 99- 5d =99-5x 13 = 99-65 =34 S3= 3/2[ 2a+2d] = 3/2[ 34+ 26] =3/2 x 60 = 90 ie 17+30+ 43=90
Answered by DOVE - Fri Sep 18 00:44:42 2009
What is the formula for the sum of an arithmetic series?
Q. this is the sequence i am working with 2,4,6,8,10
Asked by nightwing0071a - Thu Mar 9 09:19:35 2006 - - 1 Answers - 0 Comments
A. Fortunately, there is a simple formula for finding the sum of an Arithmetic Series. In general, the sum of an Arithmetic Series can be written as follows: S = i + (i+d) + (i+2d) + (i+3d) + ... +[i+(n-1)d] and can be simplified as: S = (n / 2) * [2i + (n-1) * d] where i is the initial term, d is the difference between successive terms and n is the number of terms in the series.
Answered by ineverknew - Thu Mar 9 09:24:02 2006
Q. this is the sequence i am working with 2,4,6,8,10
Asked by nightwing0071a - Thu Mar 9 09:19:35 2006 - - 1 Answers - 0 Comments
A. Fortunately, there is a simple formula for finding the sum of an Arithmetic Series. In general, the sum of an Arithmetic Series can be written as follows: S = i + (i+d) + (i+2d) + (i+3d) + ... +[i+(n-1)d] and can be simplified as: S = (n / 2) * [2i + (n-1) * d] where i is the initial term, d is the difference between successive terms and n is the number of terms in the series.
Answered by ineverknew - Thu Mar 9 09:24:02 2006
What is the formula for a sum of an arithmetic progression/series?
Q. The thing is, I've found so many different formulas, I'm not sure which one to use. Which one is for what and what is the standard one? Thanks!
Asked by Samantha - Fri Mar 28 08:24:36 2008 - - 5 Answers - 0 Comments
A. s= n/2(2a+(n-1)d) where s = sum of progression n = number of terms a = value of first term d = difference between 2 terms
Answered by hottie - Fri Mar 28 08:37:31 2008
Q. The thing is, I've found so many different formulas, I'm not sure which one to use. Which one is for what and what is the standard one? Thanks!
Asked by Samantha - Fri Mar 28 08:24:36 2008 - - 5 Answers - 0 Comments
A. s= n/2(2a+(n-1)d) where s = sum of progression n = number of terms a = value of first term d = difference between 2 terms
Answered by hottie - Fri Mar 28 08:37:31 2008
Given a1 = -23, n = 25, and an = 57, find the sum of the arithmetic series.?
Q. Given a(base)1 = -23, n = 25, and a(base)n = 57, find the sum of the arithmetic series. is this telling us to add everything together?? i don't get it...and what does arithmetic mean??
Asked by Rinoa - Sun Jul 12 13:07:22 2009 - - 1 Answers - 1 Comments
A. Sn = [ a + ( a + d) + ---a + (n - 1) d ] Sn = [ a + (n - 1) d + ---(a + d) + a ] 2 Sn = n [ 2a + (n - 1) d ] 2 Sn = n [ a + a + (n - 1) d ] Sn = ( n/2 ) [ a + { a + (n - 1) d } ] Sn = ( n/2 ) [ First + Last ] S25 = (25/2) [ - 23 + 57 ] S25 = 25 x 17 S25 = 425
Answered by como - Sun Jul 12 13:45:38 2009
Q. Given a(base)1 = -23, n = 25, and a(base)n = 57, find the sum of the arithmetic series. is this telling us to add everything together?? i don't get it...and what does arithmetic mean??
Asked by Rinoa - Sun Jul 12 13:07:22 2009 - - 1 Answers - 1 Comments
A. Sn = [ a + ( a + d) + ---a + (n - 1) d ] Sn = [ a + (n - 1) d + ---(a + d) + a ] 2 Sn = n [ 2a + (n - 1) d ] 2 Sn = n [ a + a + (n - 1) d ] Sn = ( n/2 ) [ a + { a + (n - 1) d } ] Sn = ( n/2 ) [ First + Last ] S25 = (25/2) [ - 23 + 57 ] S25 = 25 x 17 S25 = 425
Answered by como - Sun Jul 12 13:45:38 2009
Math help in Arithmetic Series please?
Q. An arithmetic series has eleven terms. The first term is 6 and the last term is (-27). Find the sum of series.
Asked by Noushy - Wed Sep 30 08:05:02 2009 - - 2 Answers - 0 Comments
Q. An arithmetic series has eleven terms. The first term is 6 and the last term is (-27). Find the sum of series.
Asked by Noushy - Wed Sep 30 08:05:02 2009 - - 2 Answers - 0 Comments
Math Homework help arithmetic series?
Q. I really need some help with this math question, all i can figure out is that its an arithmetic series: The sum of the first 10 terms of an arithmetic series is 100 and the sume of the next 10 terms is 300. find the first 3 terms of the series.
Asked by Chelsea - Thu May 21 04:45:22 2009 - - 1 Answers - 0 Comments
Q. I really need some help with this math question, all i can figure out is that its an arithmetic series: The sum of the first 10 terms of an arithmetic series is 100 and the sume of the next 10 terms is 300. find the first 3 terms of the series.
Asked by Chelsea - Thu May 21 04:45:22 2009 - - 1 Answers - 0 Comments
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