Yahoo Answers: Answers and Comments for Find the sum of the first 25 terms of an arithmetic sequence whose first term is 9 and whose common difference is 5.? [Mathematics]
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From Emem
enUS
Tue, 18 Jun 2019 10:45:41 +0000
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Yahoo Answers: Answers and Comments for Find the sum of the first 25 terms of an arithmetic sequence whose first term is 9 and whose common difference is 5.? [Mathematics]
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From Φ² = Φ+1: S_n = n*a + d*(n1)*n/2
25 * 9 + 5 * 24 * 25 ...
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https://answers.yahoo.com/question/index?qid=20190618104541AAskC38
Tue, 18 Jun 2019 10:52:25 +0000
S_n = n*a + d*(n1)*n/2
25 * 9 + 5 * 24 * 25 / 2 = 1275

From la console: For an arithmetic sequence:
a₁ =  9
a₂ = a₁ +...
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Tue, 18 Jun 2019 11:13:18 +0000
For an arithmetic sequence:
a₁ =  9
a₂ = a₁ + d ← where d is the common difference, i.e.: 5
a₃ = a₂ + d = a₁ + 2d
a₃ = a₃ + d = a₁ + 3d
…and you can generalize writing:
a(n) = a₁ + (n  1).d → for the 25th term, n = 25
a₂₅ =  9 + (25  1).5
a₂₅ =  9 + 120
a₂₅ = 111
s = a₁ + a₂ + a₃ + a₃ + … + a₂₅
s = ( 9) + ( 4) + (1) + (6) + … + (111)
s = (111) + (106) + (101) + (96) + … + ( 9)
2s = (102) * 25
s = 51 * 25
s = 1275

From Krishnamurthy: 9, 4, 1, 6, ...
an = 5 n  14
S25 = (25/2)(...
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https://answers.yahoo.com/question/index?qid=20190618104541AAskC38
Tue, 18 Jun 2019 11:07:42 +0000
9, 4, 1, 6, ...
an = 5 n  14
S25 = (25/2)(9 + 111) = 5100/4 = 1275

From Captain Matticus, LandPiratesInc: T[1] = 9 = 14 + 5 * 1
T[n] = 14 + 5 * n
T[2...
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https://answers.yahoo.com/question/index?qid=20190618104541AAskC38
Tue, 18 Jun 2019 10:53:57 +0000
T[1] = 9 = 14 + 5 * 1
T[n] = 14 + 5 * n
T[25] = 14 + 5 * 25 = 125  14 = 111
S[n] = (n/2) * (T[1] + T[n])
S[25] = (25/2) * (9 + 111)
S[25] = (25/2) * (102)
S[25] = 25 * 51
S[25] = 25 * 50 + 25 * 1
S[25] = 1250 + 25
S[25] = 1275

From Learner: Apply the formula
Sum to n terms of an AP =
(...
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Tue, 18 Jun 2019 10:53:15 +0000
Apply the formula
Sum to n terms of an AP =
(n/2)*[2a + (n  1)d]
where n = number of terms = 25
a is first term = 9
d is common difference = 5
Ans: 1275