The first four terms are 3 9 27 81 and 729 is the 6th term.
If the sequence is 1,4,10,19,31,...... Then the sequence formula is, 1 + 3/2n(n - 1) Confirm 5th term....1 + (3/2 x 5 x 4) = 1 + 30 = 31 the 6th (next) term = 1 + (3/2 x 6 x 5) = 1 + 45 = 46
Any number you like.If you fit a linear equation, Un = 24 - 11n then the 6th term is 42.However,Vn = (-29x4 + 290x3 - 1015x2 + 790x + 744)/60 gives the sequence 13, 2, -9, -20, -42.6, and -100Wn = (7x4 - 70x3 + 245x2 - 570x + 648)/20 gives the sequence 13, 2, -9, -20, -22.6, and 0whileXn = (71x4 - 710x3 + 2485x2 - 4210x + 3144)/60 gives the sequence 13, 2, -9, -20, -2.6, and 100By suitable choice of polynomial, any number at all can be in sixth place.
There are infinitely many formulae that generate that sequence for the first 5 terms, but then give different terms for the 6th and onwards. However, the simplest formula (and the one that I guess your teacher is expecting) is based on the fact that there is a fixed common difference of 6 between terms, giving t(n) = 6n - 4 for n = 1, 2, 3, ...
The question seems to be incorrect because, if the definition is 2n + 3 then the third tern should be 9 and the fourth one 11. In that case, t(6) = 2*6 + 3 = 12+3 = 15
If you mean 6, 18, 54 then the 6th term is 1456 because each term is 3 times greater than the previous term
the answer is 8
To find the first term and common ratio of a geometric progression, we can use the formula for the nth term of a geometric sequence: (a_n = a_1 \times r^{(n-1)}). Given that the 6th term is 160 and the 9th term is 1280, we can set up two equations using these values. From the 6th term, we get (a_1 \times r^5 = 160), and from the 9th term, we get (a_1 \times r^8 = 1280). By dividing the two equations, we can eliminate (a_1) and solve for the common ratio (r).
The first four terms are 3 9 27 81 and 729 is the 6th term.
It appears that a number of -79 is missing in the sequence and so if you meant -58 -65 -72 -79 -86 then the nth term is -7n-51 which makes 6th term in the sequence -93
If the sequence is 1,4,10,19,31,...... Then the sequence formula is, 1 + 3/2n(n - 1) Confirm 5th term....1 + (3/2 x 5 x 4) = 1 + 30 = 31 the 6th (next) term = 1 + (3/2 x 6 x 5) = 1 + 45 = 46
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 this sequence, we first need to identify the pattern. The differences between consecutive terms are 12, 20, 28, 36, and 44. These are increasing by 8 each time. This means the second difference is constant, indicating a quadratic sequence. By calculating the second difference, we can determine the equation for the nth term. The nth term for this sequence is n^2 + 10.
It is x6.
Any number you like.If you fit a linear equation, Un = 24 - 11n then the 6th term is 42.However,Vn = (-29x4 + 290x3 - 1015x2 + 790x + 744)/60 gives the sequence 13, 2, -9, -20, -42.6, and -100Wn = (7x4 - 70x3 + 245x2 - 570x + 648)/20 gives the sequence 13, 2, -9, -20, -22.6, and 0whileXn = (71x4 - 710x3 + 2485x2 - 4210x + 3144)/60 gives the sequence 13, 2, -9, -20, -2.6, and 100By suitable choice of polynomial, any number at all can be in sixth place.
The 6th number of the Fibonacci sequence is 8.0 + 0 = 00 + 1 = 11 + 1 = 21 + 2 = 32 + 3 = 53 + 5 = 8Notice how it is the 6th equation meaning its the 6th Fibonacci number.Note that some people like to use 1 twice instead of 0.http://en.wikipedia.org/wiki/Fibonacci_number
There are infinitely many formulae that generate that sequence for the first 5 terms, but then give different terms for the 6th and onwards. However, the simplest formula (and the one that I guess your teacher is expecting) is based on the fact that there is a fixed common difference of 6 between terms, giving t(n) = 6n - 4 for n = 1, 2, 3, ...