Q: What is the nth term of 3 6 11 18 27?

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The nth term of the sequence is (n + 1)2 + 2.

each time the number increases by 8 and the original number was 3.This means that nth term = 3+(n-1)8ie. the fourth term 27 would be 3 + (4-1)x8 = 3+24 = 27

2n +29

(n(n-1) divided by 2 + 1) multiplied by 3) +6

n3

Willies

5n+2

There is no pattern

9 3 1

There are many possible answers, but the simplest is t(n) = 27 - 8*n where n = 1, 2, 3, ...

The pattern for the nth term is n3. Therefore, the fourth term in the sequence is equal to 43 = 64.

3^n These are powers of 3

the anser is that you are stupid

I can tell you that it is increasing 9, 15, 21, 27. i.e. adding on 6 to the gap each time so if you mean the next term by nth term, then it will be adding on 33, which will leave you at 113

ans=102 HOW? 6,11,13,27,38,.... 6(+5),11(+7),13(+9),27(+11),38(+13),51(+15),66(+17),83(+19),102adding 5 to 6 gives 11 adding 7(two greater than previous one) gives 13...and so on... to find the nth term N squared +2

11/18 × 27/50 = 33/100

It is: 27-2n

27/18 = 3/2 = 11/2

Subtract 2 from the sequence and the answer becomes obvious: 1, 4, 9, 16, 25,...,N2 Now, add the 2 back in: 3, 6, 11, 18, 27,...,(N2+2)

nth term is 8 - n. an = 8 - n, so the sequence is {7, 6, 5, 4, 3, 2,...} (this is a decreasing sequence since the successor term is smaller than the nth term). So, the sum of first six terms of the sequence is 27.

A single number, such as 1521273339 does not define a sequence. There is no nth term for a signle number.

So if we call the first term of the sequence A,and the common difference d, then the formula for the nth term is A+ [(n-1)d] this just comes from understanding that the first term is A the second term is A+d the third is A+2d the 4th is A+ 3d so the nth term is A + (n-1)d. So let's set up two equations and solve them for A and d and then we are done A+ 8d=90 A+15d=153 This is the same as -A-8d=-90 A+15d=153 and we add these two equation to eliminate A. This is called elimination. You can also solve this system of 2 equations in 2 unknowns with substitution by substituting A=90-8d into the second equation. I think this is harder We have 7d=63 so d=9 and then plug that into either equation and A=18 So the first 4 terms are 18, 18+9, 18+18, 18+27 or we write 18, 27, 36, 45

If by using the formula: nth term = n2+2 the sequence is 3 6 11 18 27 38 51 66.Then it follows that by using the same formula the sequence is III VI XI XVIII XXVII XXXVIII LI LXVI in Roman numerals.

after -9 it is -15 then -21, -27 and the ninth is -36

tn = 3n or 3*3* ... *3 (where there are n threes.)

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