Q: What is the nth term for 11 21 31 41?

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31 - n

The nth term is 6n+1 and so the next term will be 31

t(n) = 4*n2 - 5

The nth term in the arithmetic progression 10, 17, 25, 31, 38... will be equal to 7n + 3.

+9

Un = (-n5 + 20n4 - 155n3 + 820n2 - 1044n + 420)/60

The pattern is: +11, +15, +19, +23, +27 (4n+7) So, the next number would be: (4*6 + 7) = 24 + 7 = +31 Therefore, the answer is: 105 + 31 = 136

8n + 7

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

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

2n^2-1

The nth term is 7n-4 and so the next number in the sequence is 31

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

t(n) = 3n2 + n + 1

The sequence progresses by adding 7 to the previous term.The nth term is thus equal to 10 + 7n. The 11th term therefore is equal to 10 + (7 * 11) = 10 + 77 = 87.

tn = 1.5*n2 - 1.5*n + 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

Yes. Each term is 10 more than the previous term.

a (sub n) = 35 - (n - 1) x d

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.

31

Subtract 7 from each number, so the 9th number would be 59.

The differnace between the numbr is 7, therefore the first part of the formula will be 7n, now for the first term we replace n with one so nx7=7 and to get to 10 you need to add 3 making the nth term 7n+3, To check your answer you must replace n with the two (for the second term) which comes to 14 to get to seventeen you need to add three so the formula nth term 7n+3, hope this helped

You can see that all the numbers go up by 7. This means that the first part of the nth term rule for this sequence is 7n. Now, you have to find out how to get from 7 to 3, 14 to 10, 21 to 17 ... this is because we are going up in the 7 times table. To get from the seventh times table to the sequence, you take away four. So the answer is : 7n-4

11 and 31 and all the fives between numbers.