To find the nth term of the sequence -4, -1, 4, 11, 20, 31, we first identify the pattern in the differences between the terms: 3, 5, 7, 9, 11, which increases by 2 each time. This suggests a quadratic relationship. The nth term can be expressed as ( a_n = n^2 + n - 4 ). Thus, the nth term of the sequence is ( a_n = n^2 + n - 4 ).
The given sequence is 11, 31, 51, 72 The nth term of this sequence can be expressed as an = 11 + (n - 1) × 20 Therefore, the nth term is 11 + (n - 1) × 20, where n is the position of the term in the sequence.
To find the nth term formula for the sequence -4, -1, 4, 11, 20, 31, we first observe the differences between consecutive terms: 3, 5, 7, 9, 11, which are increasing by 2. This indicates a quadratic relationship. The nth term formula can be derived as ( a_n = n^2 + n - 4 ).
tn = 1.5*n2 - 1.5*n + 1
The sequence given is an arithmetic sequence where each term increases by 6. To find the nth term, you can use the formula for the nth term of an arithmetic sequence: ( a_n = a_1 + (n-1)d ), where ( a_1 ) is the first term and ( d ) is the common difference. Here, ( a_1 = 7 ) and ( d = 6 ). Thus, the nth term can be expressed as ( a_n = 7 + (n-1) \times 6 = 6n + 1 ).
Assuming this is a linear or arithmetic sequence, the nth term is Un = 31 - 8n. But, there are infinitely many polynomials of order 5 or higher, and many other functions that will fit the above 5 numbers.
The given sequence is 11, 31, 51, 72 The nth term of this sequence can be expressed as an = 11 + (n - 1) × 20 Therefore, the nth term is 11 + (n - 1) × 20, where n is the position of the term in the sequence.
To find the nth term formula for the sequence -4, -1, 4, 11, 20, 31, we first observe the differences between consecutive terms: 3, 5, 7, 9, 11, which are increasing by 2. This indicates a quadratic relationship. The nth term formula can be derived as ( a_n = n^2 + n - 4 ).
+9
Difference is 5,7,9,11,13 Second difference is 2 (2x)^2 gives 4,9,16,25 Difference between 2x^2 and sequence is -5 Thus, the nth term will be (2n)^2-5
2n^2-1
31 - n
10n + 1
The nth term in the arithmetic progression 10, 17, 25, 31, 38... will be equal to 7n + 3.
The nth term is 6n+1 and so the next term will be 31
the nth term is = 31 + (n x -9) where n = 1,2,3,4,5 ......... so the 1st term is 31+ (1x -9) = 31 - 9 =22 so the 6th tern is 31 + (6 x -9) = -23 Hope this helps
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