Q: What is first four terms of the arithmetic sequence with common difference of 3 and a first term of -26?

Write your answer...

Submit

Related questions

6

16

It is a + 8d where a is the first term and d is the common difference.

What is the 14th term in the arithmetic sequence in which the first is 100 and the common difference is -4? a14= a + 13d = 100 + 13(-4) = 48

From any term after the first, subtract the preceding term.

6

A single number, such as 11111, cannot define an arithmetic sequence. On the other hand, it can be the first element of any kind of sequence. On the other hand, if the question was about ``1, 1, 1, 1, 1'' then that is an arithmetic sequence as there is a common difference of 0 between each term.

You subtract any two adjacent numbers in the sequence. For example, in the sequence (1, 4, 7, 10, ...), you can subtract 4 - 1, or 7 - 4, or 10 - 7; in any case you will get 3, which is the common difference.

You take the difference between the second and first numbers.Then take the difference between the third and second numbers. If that difference is not the same then it is not an arithmetic sequence, otherwise it could be.Take the difference between the fourth and third second numbers. If that difference is not the same then it is not an arithmetic sequence, otherwise it could be.Keep checking until you think the differences are all the same.That being the case it is an arithmetic sequence.If you have a position to value rule that is linear then it is an arithmetic sequence.

It is an Arithmetic Progression with a constant difference of 11 and first term 15.

The sequence in the question is NOT an arithmetic sequence. In an arithmetic sequence the difference between each term and its predecessor (the term immediately before) is a constant - including the sign. It is not enough for the difference between two successive terms (in any order) to remain constant. In the above sequence, the difference is -7 for the first two intervals and then changes to +7.

6

The sum of the first 12 terms of an arithmetic sequence is: sum = (n/2)(2a + (n - 1)d) = (12/2)(2a + (12 - 1)d) = 6(2a + 11d) = 12a + 66d where a is the first term and d is the common difference.

Since there are no graphs following, the answer is none of them.

a + 99d where 'a' is the first term of the sequence and 'd' is the common difference.

35 minus 4 differences, ie 4 x 6 so first term is 11 and progression runs 11,17,23,29,35...

An arithmetic sequence is a group or sequence of numbers where, except for the first number, each of the subsequent number is determined by the same rule or set of rules. * * * * * The above answer is incorrect. The rule can only be additive: it cannot be multiplicative or anything else.

The graph will be a set of disjoint points with coordinates [n, 0.5*(1+n)]

The difference between successive terms in an arithmetic sequence is a constant. Denote this by r. Suppose the first term is a. Then the nth term, of the sequence is given by t(n) = (a-r) + n*r or a + (n-1)*r

An arithmetic sequence is a sequence of numbers such that the difference between successive terms is a constant. This constant is called the common difference and is usually denoted by d. If the first term is a, then the iterative definition of the sequence is U(1) = a, and U(n+1) = U(n) + d for n = 1, 2, 3, ... Equivalently, the position-to-term rule which defines the sequence is U(n) = a + (n-1)*d for n = 1, 2, 3, ...

An arithmetic sequence depends on two numbers: the seed, a, and the common difference, d.The first number in the sequence is the seed and each number is obtained from the previous one by adding the difference.So U1 = aU2 = a + dU3 = U2 + d = a + d + d = a + 2dU4 = U3 + d = a + 2d + d = a + 3dand so on.In general, Un = a + (n - 1)*d

In this case you can calculate the 14th. term as 100 + (-4)(14-1).

tn = a + (n - 1)d where a is the first term and d is the difference between each term.

This is an arithmetic sequence with the first term t1 = 1, and the common difference d = 6. So we can use the formula of finding the nth term of an arithmetic sequence, tn = t1 + (n - 1)d, to find the required 30th term. tn = t1 + (n - 1)d t30 = 1 + (30 - 1)6 = 175

First we define an arithmetic sequence as one where each successive term has a common difference and that difference is constant. An example might be 1, 4, 7, 10, 13, 16, ..where the difference is 3. 1+3=4, 4+3=7 etc. Here is a common example that is given as a problem but shows a real life example of arithmetic sequences. A theater has 60 seats in the first row, 68 seats in the second row, 76 seats in the third row, and so on in the same increasing pattern. If the theater has 20 rows of seats, how many seats are in the theater? The common difference is 8 and we want the the sum of the first 20 terms this gives us the sum of all the seats. We solve this by first finding the 20th term which is 212 and noting that the first term is 60. We add the first and the 20th terms in the sequence and multiply the sum by 20. Next we divide that product by 2. The sum we are looking for is 20(60+212)/2=2720 so there are 2720 seats in the theater! The general formula to find the sum of the first n terms in an arithmetic sequence is to multiply n by the sum of the first and nth terms in the sequence and divide that answer by 2. In symbols we write Sn=n(a1+ an)/2

Study guides