The nth term of the sequence is 3n-8 and so the 30th term is 3*30 -8 = 82
The given sequence is a geometric sequence where each term is multiplied by 2 to get the next term. The first term (a) is 4, and the common ratio (r) is 2. The nth term of a geometric sequence can be found using the formula ( a_n = a \cdot r^{(n-1)} ). Therefore, the nth term of this sequence is ( 4 \cdot 2^{(n-1)} ).
The sequence 1, 3, 5, 7, 9 is an arithmetic sequence where each term increases by 2. The nth term can be expressed as ( a_n = 2n - 1 ). Therefore, for any positive integer ( n ), the nth term of the sequence is ( 2n - 1 ).
1 - 2 - 4 - 8 - 16 - 32 - 64 the sequence doubles
-1
Triangular numbers are numbers in the sequence 1, 1+2, 1+2+3, 1+2+3+4. This sequence can be represented by triangles as follows: (very crude figure with an even cruder browser!)xxxxxxxxxxxxxxxxxxxxand so on.The nth term of this sequence is n*(n+1)/2.Triangular numbers are numbers in the sequence 1, 1+2, 1+2+3, 1+2+3+4. This sequence can be represented by triangles as follows: (very crude figure with an even cruder browser!)xxxxxxxxxxxxxxxxxxxxand so on.The nth term of this sequence is n*(n+1)/2.
82
The general term for the sequence 0, 1, 1, 2, 2, 3, 3 is infinite sequence.
The nth term of the sequence is 3n - 2.
The given sequence is a geometric sequence where each term is multiplied by 2 to get the next term. The first term (a) is 4, and the common ratio (r) is 2. The nth term of a geometric sequence can be found using the formula ( a_n = a \cdot r^{(n-1)} ). Therefore, the nth term of this sequence is ( 4 \cdot 2^{(n-1)} ).
The nth term of the sequence 2n + 1 is calculated by substituting n with the term number. So, the tenth term would be 2(10) + 1 = 20 + 1 = 21. Therefore, the tenth term of the sequence 2n + 1 is 21.
1 - 2 - 4 - 8 - 16 - 32 - 64 the sequence doubles
-1
20th term = 20*(20+1)/2
Triangular numbers are numbers in the sequence 1, 1+2, 1+2+3, 1+2+3+4. This sequence can be represented by triangles as follows: (very crude figure with an even cruder browser!)xxxxxxxxxxxxxxxxxxxxand so on.The nth term of this sequence is n*(n+1)/2.Triangular numbers are numbers in the sequence 1, 1+2, 1+2+3, 1+2+3+4. This sequence can be represented by triangles as follows: (very crude figure with an even cruder browser!)xxxxxxxxxxxxxxxxxxxxand so on.The nth term of this sequence is n*(n+1)/2.
the (n-1)th term plus the (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).
Each number in this sequence is twice the previous number. The nth. term is 2n-1.Each number in this sequence is twice the previous number. The nth. term is 2n-1.Each number in this sequence is twice the previous number. The nth. term is 2n-1.Each number in this sequence is twice the previous number. The nth. term is 2n-1.