find the sum of 5+8+11+....to10 terms using the formula
Answers
Answer:
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Step-by-step explanation:
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Answer:
Answer: 25.
ns(n)t(n)1025=1,2,3,…=2+3n=5,8,11,…=∑k=1n2+3∑k=1nn=2n+3n(n+1)2=3n2+7n2=5,13,24,…=3n2+7n2⇒n=25
>>> [(n,2+3*n) for n in xrange(1,26)]
[(1, 5), (2, 8), (3, 11), (4, 14), (5, 17), (6, 20), (7, 23), (8, 26), (9, 29), (10, 32), (11, 35), (12, 38), (13, 41), (14, 44), (15, 47), (16, 50), (17, 53), (18,
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the series 5+8+11+….. can be written as 5+8+11+…+3n+2
the sum of the series can be written as S=5+8+11+…+3n+2. Also S=(3n+2)+(3n-1)+…+5. Therefore we can pair each number to each other such that each pair totals (3n+2)+5 or 3n+7. Therefore 2S=(3n+7) n times, that is n(3n+7) or 2S=n(3n+7)
2S=n(3n+7)
2S=3n²+7n since we want S =1025 2S=2050
2050=3n²+7n
3n²+7n-2050=0 then using using the quadratic formula for a=3, b=7, and c=-2050 we get n=(-7+/-157)/6. Since only the positive root will satisfy our question we take n=(-7+157)/6
n=25
the 25th term will be 3n+2 for n=25 . therefore the 25th term will be 3x25+2=77
Proof: S=(1/2)*n*(3n+7) set n=3 and solve to get S=1025
therefore there are 25 terms in the series when the total of the series is 1025.