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.
tn = 1.5*n2 - 1.5*n + 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.
1 and 11/20
There are infinitely many possible answers. the simplest polynomial is Un = -n4 + 12n3 - 47n2 + 78n - 41
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.
+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
63
11
5 to 7 is 27 to 17 is 1017 to 19 is 219 to 29 is 1029 to 31 is 2there fore following the pattern the nth term is 4131 to 41 is 10