This is an Arithmetic Progression or Sequence with an Initial Term (a) of 10 and a Common Difference (d) of -3
The standard formula is U(n) = a + (n - 1)d
Substituting the values for a and d gives :-
U(n) = 10 + [(n - 1) x (-3)] = 10 - 3n + 3
We can write this as, 13 - 3n but it is common practice to retain the initial term (10)............so, U(n) = 10 - 3(n - 1)
Then U(5) = 10 - 3(5 - 1) = 10 - 12 = -2 which confirms the value shown in the question.
If the term number is n, then the nth term is 10(n-1) +8.
The nth term is 4n - 3
10n + 1
It could be 3691. Use the position to term rule: t(n) = (856n3 - 5091n2 + 9545n - 5304)/6 for n = 1, 2, 3, ...
PRIME
If the term number is n, then the nth term is 10(n-1) +8.
The nth term is 4n - 3
1 2 3 4 5 2 5 8 11 14 ... If this is the sequence, the position-to-term rule is 3n-1. However, it could be another sequence depending on the rest of the terms.
term n = 3n - 1 for n = 1, 2, 3, ...
t(n) = 50 - 2n where n = 1, 2, 3, ...
To find the first, fourth, and tenth terms of the arithmetic sequence defined by the rule ( A(n) = 1 + (n - 1)(-4.1) ), we can substitute ( n ) with 1, 4, and 10. For ( n = 1 ): ( A(1) = 1 + (1 - 1)(-4.1) = 1 ). For ( n = 4 ): ( A(4) = 1 + (4 - 1)(-4.1) = 1 - 12.3 = -11.3 ). For ( n = 10 ): ( A(10) = 1 + (10 - 1)(-4.1) = 1 - 36.9 = -35.9 ). Thus, the first term is 1, the fourth term is -11.3, and the tenth term is -35.9.
1, 4, 7, 10, 13, …
Term to Term rule in Maths is how much you go up or down in. e.g 1,2,3,4,5,6 would be +1
10n + 1
The rule is multiply the previous term by -1 to find the next term.
The pattern consists of the cubes of consecutive integers. Specifically, the numbers are (1^3), (2^3), (3^3), (4^3), and (5^3), resulting in 1, 8, 27, 64, and 125, respectively. The rule for this pattern is that each term is equal to (n^3), where (n) is the position of the term in the sequence (starting from 1).
It could be 3691. Use the position to term rule: t(n) = (856n3 - 5091n2 + 9545n - 5304)/6 for n = 1, 2, 3, ...