Well, isn't that just a happy little sequence we have here! To find the pattern, we can see that the differences between the terms are increasing by 2 each time. So, the nth term can be found by the formula n^2 + 4. Just like painting a beautiful landscape, sometimes all we need is a little patience and observation to uncover the hidden beauty within numbers.
Well, darling, the nth term for this sequence is 8n + 7. You just add 8 to each term to get the next one, simple as that. So, if you want the 100th term, just plug in n=100 and you'll get 807. Easy peasy lemon squeezy!
The nth term is 7n-4 and so the next number in the sequence is 31
Type your answer here... The next numbers in the sequence are 55, 70, 87, 106, 127, etc.
10n + 1
As given, the sequence is too short to establish the generating rule. If the second term was 19 and NOT 29, then the nth term is tn = 6*n + 7 or 6(n+1)+1
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.
The nth term in the arithmetic progression 10, 17, 25, 31, 38... will be equal to 7n + 3.
Well, darling, the nth term for this sequence is 8n + 7. You just add 8 to each term to get the next one, simple as that. So, if you want the 100th term, just plug in n=100 and you'll get 807. Easy peasy lemon squeezy!
The nth term is 7n-4 and so the next number in the sequence is 31
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.
Type your answer here... The next numbers in the sequence are 55, 70, 87, 106, 127, etc.
10n + 1
As given, the sequence is too short to establish the generating rule. If the second term was 19 and NOT 29, then the nth term is tn = 6*n + 7 or 6(n+1)+1
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
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 ).
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
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