That depends what the pattern of the sequence is.
taking the 1st term as n=1 the 2nd term as n= 2 etc then the first 5 terms of 16-3n are : 1st : 16-3 = 13 2nd : 16-6 = 10 3rd : 16-9 = 7 4th: 16-12 = 4 5th : 16-15 = 1 hope this helps
Nth number in an arithmetic series equals 'a + nd', where 'a' is the first number, 'n' signifies the Nth number and d is the amount by which each term in the series is incremented. For the 5th term it would be a + 5d
It will have to be a very complex power function because the are not increasing steadily. The increase between the 5th and 6th term is smaller than that between the 4th and 5th terms.
quinquennial
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earth is 5th number in term of size
That depends what the pattern of the sequence is.
taking the 1st term as n=1 the 2nd term as n= 2 etc then the first 5 terms of 16-3n are : 1st : 16-3 = 13 2nd : 16-6 = 10 3rd : 16-9 = 7 4th: 16-12 = 4 5th : 16-15 = 1 hope this helps
Richard J. Daley served the following: Terms of Office 1st term: 1955-1959 2nd term: 1959-1963 3rd term: 1963-1967 4th term:1967-1971 5th term: 1971-1975 6th term: 1975-1976 (died in office) see also, Chicago's Mayors, Chicago Public Library at: ChiPubLib
Nth number in an arithmetic series equals 'a + nd', where 'a' is the first number, 'n' signifies the Nth number and d is the amount by which each term in the series is incremented. For the 5th term it would be a + 5d
FDR was the only president to server more than 2 terms so it must be him, Franklin died of polio at the begining of his 4th or 5th term i don't remember which
2500, 100n2Restate the question: what are the 5th and nth term of (10n)2?If this is not your question, please clarify and resubmit the question.Assuming the first term is when n=1, then the 5th term is (10x5)2 = (50)2 =2500.The nth term would be just (10n)2, although you could expand and simplify to get (102)(n2) = 100n2.
The series given is an arithmetic progression consisting of 5 terms with a common difference of 5 and first term 5 → sum{n} = (n/2)(2×5 + (n - 1)×5) = n(5n + 5)/2 = 5n(n + 1)/2 As no terms have been given beyond the 5th term, and the series is not stated to be an arithmetic progression, the above formula only holds for n = 1, 2, ..., 5.
I'll try to answer the question, "If the 5th term of a geometric progression is 2, then the product of its FIRST 9 terms is --?" Given the first term is A and the ratio is r, then the progression starts out... A, Ar, Ar^2, Ar^3, Ar^4, ... So the 5th term is Ar^4, which equals 2. The series continues... Ar^5, Ar^6, Ar^7, Ar^8, ... Ar^8 is the 9th term. The product P of all 9 terms is therefore: P = A * Ar * Ar^2 *...*Ar^8 Collect all the A's P = (A^9)*(1 * r * r^2 ...* r^8) P = A^9 * r^(0+1+2+...+8) There's a formula for the sum of the first n integers (n/2)(n+1), or if you don't know just add it up. 1+2+...+8 = 36 Therefore P = A^9 * r^36 Since 36 is a multiple of 9, you can simplify: P = (Ar^4)^9 Still with me? Remember that Ar^4=2 (a given fact). So finally P = 2^9 = 512. Cute problem.
Each term seems to be double of the previous number starting with 3. Hence 4th term = 24 and 5th is 48
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