u15 = u7 + (15-7)*d = 2.33 + 8*(-0.67) = -3.03
The nth term for that arithmetic progression is 4n-1. Therefore the next term (the fifth) in the sequence would be (4x5)-1 = 19.
There are two ways to say the general rule. They both mean exactlythe same thing, and they both generate the same sequence:1). Starting with 15, each new term is 3 less than the one before it.2). The nth term of the sequence is [ 18 - 3n ] or [ 3 times (6 - n) ].
The next term is: 15+4 = 19 The nth term is: 4n-5
Suppose the sequence is defined by an = a0 + n*d Then a1 = a0 + d = 15 and a13 = a0 + 13d = -57 Subtracting the first from the second: 12d = -72 so that d = -6 and then a0 - 6 = 15 gives a0 = 21 So a32 = 21 - 32*6 = -171
The nth term in this sequence is 4n + 3.
It is an Arithmetic Progression with a constant difference of 11 and first term 15.
The nth term is -7n+29 and so the next term will be -6
The nth term for that arithmetic progression is 4n-1. Therefore the next term (the fifth) in the sequence would be (4x5)-1 = 19.
yes it is
There are two ways to say the general rule. They both mean exactlythe same thing, and they both generate the same sequence:1). Starting with 15, each new term is 3 less than the one before it.2). The nth term of the sequence is [ 18 - 3n ] or [ 3 times (6 - n) ].
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.
The next term is: 15+4 = 19 The nth term is: 4n-5
That is called an arithmetic sequence. For example: 8, 15, 22, 29, 36, 43, 50, 57, etc.
Suppose the sequence is defined by an = a0 + n*d Then a1 = a0 + d = 15 and a13 = a0 + 13d = -57 Subtracting the first from the second: 12d = -72 so that d = -6 and then a0 - 6 = 15 gives a0 = 21 So a32 = 21 - 32*6 = -171
An Arithmetic Sequence can be written as :-a, (a + d), (a + 2d), . . ., [a + (n - 1)d], where a is the first term and d is the common difference.So, if a(3) = a + 2d = 1985and, a(8) = a + 7d = 2015Then subtracting the first equation from the second gives :-5d = 30 : d = 6 . . . and replacing this value in either equation (let's take the first)a + 2d = a +2*6 = a + 12 = 1985, a = 1973.Thus, a(15) = 1973 + (14*6) = 1973 + 84 = 2057.Another Answer:-The nth term is 6n+1967 and the 15th term is 2057
For an Arithmetic Progression, Sum = 15[a + 7d].{a = first term and d = common difference} For a Geometric Progression, Sum = a[1-r^15]/(r-1).{r = common ratio }.
The nth term in this sequence is 4n + 3.