Q&A for How to Find a Number of Terms in an Arithmetic Sequence

As explained above, n = [(L - A) / d] + 1.

Subtract any term from the one that follows it.

You can't without knowing the difference between consecutive terms.

Here is an example: -2, -5, -8, -11. In this sequence the common difference (d) is found by subtracting any term from the term that precedes it. For instance, -5 - (-2) = -5 + 2 = -3. The common difference is a negative 3, which means 3 is always subtracted from a given term to find the next term in the sequence.

When given a sequence, you'd usually get the first term. Unless n(1)=0, then no, the arithmetic sequence doesn't always begin with 0.

To find the number of terms in an arithmetic sequence, divide the common difference into the difference between the last and first terms, and then add 1.

You would need more information, such as the common difference and the first and last terms.

Assuming this sequence consists of every integer from 31 through 49, subtract 31 from 49, then add 1.

Finding the sum of an arithmetic sequence involves finding the average of the first and last numbers of the sequence. Therefore, you must know the 40th term. Once you find the 40th term (there's a wikiHow article on finding a certain term in an arithmetic sequence), add it to 2, divide by 2, then multiply by 40. That's the sum you're looking for.

You don't have enough information to find the number of terms quickly. However, you could start at the first term and keep adding the common difference over and over until you reach the given sum of the sequence. Count the number of times you added the common difference. Add 1 to that number to get the number of terms in the sequence.

In a geometric progression, the ratio of any two successive terms taken in a definite order is the same wherever these two terms occur in the progression.

You can set up equations for both, but if exponential, it might go to quadratic, cubic, etc.

You would have done it wrong, so check your work. If the formula turned out to be a quadratic, then look for the positive value, and if it was not a quadratic, then check your work as I told you before.

There is no last number in an infinite sequence; by definition, it keeps going on forever.

You need to know the first, second and last terms to know the number of terms in an arithmetic sequence.

Subtract a number from the arithmetic sequence with the the number that is before it.

Use the formula tn = a + (n - 1) d to solve for n. Plug in the last term (tn), the first term (a), and the common difference (d). Work through the equation until you’ve solved for n.

A1 is the first term, so n will have the first term included in it. N is the value of how many terms there are. Since you are already using A1, you have to subtract that from n. A1 is one term, so n-1.

With the last term given, n = [ l + a]/d + 1 Where l is the last term, a is the first term, d is the common difference.

There are 16 terms. d=8, so you can divide 120 (the difference) by 8 and then add 1.

Then it's not an arithmetic sequence. Arithmetic sequences only work when the differences are the same.

You don't. You'd also have to know how many terms there are in the sequence.

You are given the first and last (sixth) terms and that n=6. Using the above formula, 136 = 36 + d(6-1) = 36 + 5d. Then 100 = 5d, and d=20. So the sequence is 36, 56, 76, 96,116, 136.

Because that's not an arithmetic sequence (since the two differences are not the same), you can't find the next two terms.

Perhaps you've been given faulty information. When an arithmetic sequence is constructed correctly, the difference between the first and last terms will be divisible by the common difference.

By inspection, we can see that there are 15 terms. We can prove it by using the formula for finding n: the nth term = 150 = the first term + (n-1)d. So 150 = 10 + (n-1)(10). Then 140 = (n-1)(10). Then 14 = (n-1), and 15 = n.