Each number is twice the previous number.
the (n-1)th term plus the (n-2)th term.
This is the Fibonacci sequence, where the number is the sum of the two preceding numbers. The nth term is the (n-1)th term added to (n-2)th term
Let n (i) = the term number of each term in the sequence., with (i) going from 1-6 E.g term number 1 (n (1) ) is 3. n(2)= -7 etc... Therefore n(i) for odd terms in the sequence is n (i)= (n (i -2)th term +1). For even terms in the sequence, n(i)= (n (i - 2)th term -3).
The nth term of a arithmetic sequence is given by: a{n} = a{1} + (n - 1)d → a{5} = a{1} + (5 - 1) × 3 → a{5} = 4 + 4 × 3 = 16.
1. Each term is half the previous term.
the (n-1)th term plus the (n-2)th term.
The 19th term of the sequence is 16.
This is the Fibonacci sequence, where the number is the sum of the two preceding numbers. The nth term is the (n-1)th term added to (n-2)th term
Let n (i) = the term number of each term in the sequence., with (i) going from 1-6 E.g term number 1 (n (1) ) is 3. n(2)= -7 etc... Therefore n(i) for odd terms in the sequence is n (i)= (n (i -2)th term +1). For even terms in the sequence, n(i)= (n (i - 2)th term -3).
The nth term of a arithmetic sequence is given by: a{n} = a{1} + (n - 1)d → a{5} = a{1} + (5 - 1) × 3 → a{5} = 4 + 4 × 3 = 16.
n2
To find the nth term in this sequence, we first need to determine the pattern. The differences between consecutive terms are 5, 7, 9, and 11 respectively. These differences are increasing by 2 each time. This indicates that the sequence is following a quadratic pattern. The nth term for this sequence can be found using the formula for the nth term of a quadratic sequence, which is Tn = an^2 + bn + c.
Well, darling, the sequence you've got there is just the perfect squares of numbers. The 8th term would be the square of the 8th number, which is 64. So, the 8th term of the sequence 1, 4, 9, 16, 25 is 64. Keep those brain cells sharp, honey!
1. Each term is half the previous term.
1 - 2 - 4 - 8 - 16 - 32 - 64 the sequence doubles
1, 16, 81, 256 14641 is the 11th term.
the series can be 1,-4,16,-64