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Find the 10th term of the series 1+2+3+4........
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10.
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John wants to calculate the sum of a geometric series with 10 terms, where the 10th term is 5 859 37... ... series with 10
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find the h.p. whose 6th term is 1 and 10th term is 5. find 15th term of the series also.
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so term(6) = 1/(a+5d) = 1 a+5d = 1 ** term(10) = 1/(a+9d) = 5 5a + 45d = 1 *** ** times 5 ---> 5a +
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the 10th term is 34 of arithmetic series and the sum of the 20th term is 710.what is the 25th term?
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the difference between T10 and T20 is 10d. So, 10d=710-34=676 d=67.6 T25=T20+5d = 710+338=1048
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find the sum of the series of (-2)^n/3^n+1.
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To determine if the series is a geometric series, we need to find the common ratio, r. We can do
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What is the sum of the last 7terms of the geometric series with 𝑡3 = 3, 𝑡7 = 768 and the last term being the 10th term?
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wrong again t7/t3 = r^4, not r^6 so, r^4 = 768/3 = 256 so r = ±4 now, t1 = 3/4^2 = 3/16 S3 = 3/16
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the 10th term of an arithmetic series is 34, and the sum of the first 20 terms is 710, determine the 25th term.
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I will assume that you know the formulas for term(n) and sum(n) of an AS 10 term is 34 ---> a+9d =
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The 10th term of an arithmetic series is 34, and the sum of the first 20 terms is 710. Determine the 25th term.
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Use your formulas a+9d = 34 (#1) (20/2)[2a + 19d] = 710 2a + 19d = 71 (#2) solve the two equations.
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The 10th term of an arithmetic series is 34, and the sum of the first 20 terms is 710. Determine the 25th term.
Can someone
Top answer:
You know t(10) = a+9d a+9d = 34 or a = 34-9d S(20) = 20/2[2a + 19d] 710 = 10(2a + 19d) 71 = 2a + 19d
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