# Video: Using Arithmetic Sequences in a Real-World Context

Mrs. Jones has \$3250 in her bank account. If she plans to deposit \$100 per month, write the first 6 terms of a sequence that would represent her monthly account balances starting with her current balance as the first term.

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### Video Transcript

Mrs. Jones has 3250 dollars in her bank account. If she plans to deposit 100 dollars per month, write the first six terms of a sequence that would represent her monthly account balances, starting with her current balance as the first term.

So we’re looking for the first six terms of the sequence that represent the amount of money Mrs. Jones has in her account each month. That’s her monthly account balances. We’re told that we should use her current balance as the first term in this sequence. And from the information given in the question, we know that her current balance is 3250 dollars. So we can go ahead and write this as the first term in our sequence.

We’re then told that Mrs. Jones plans to deposit 100 dollars per month. So every month, her account balance will be increasing by 100 dollars. This means that, to find the next term in our sequence, each time we just need to add 100 to the previous term. We’re adding 100 to the balance the month before. 3250 plus 100 is 3350. Adding 100 again gives 3450. And we can continue in this way to find the remaining three terms: 3550, 3650, and 3750.

Now this is a reason to be a straightforward problem. But there is actually another method that we could use to calculate the terms in this sequence. As the amount that we’re adding each time is constant, our sequence has what’s called a common difference. And it’s, therefore, an arithmetic sequence.

There is a general formula that we can use to find any term in an arithmetic sequence, if we know the first term and the common difference. It’s this. 𝑇𝑛 — that’s the 𝑛th term — is equal to 𝑎, the first term, plus 𝑛 minus one multiplied by 𝑑, the common difference. In our sequence, the first term is Mrs. Jones’s starting account balance. That’s 3250. And the common difference is 100. So we can use this formula to find any term in the sequence.

So, for example, to find the sixth term in the sequence, we substitute 𝑛 equals six, giving that the sixth term is equal to 3250 plus six minus one multiplied by 100. That’s 3250 plus 500, which is 3750. And if we look at the sequence we’ve already written down, we can see that this is indeed the sixth term. There wasn’t any particular need to use this more formal method in this problem. But it would be useful if we’d been asked to find, for example, the hundredth term, as we could do this without calculating all of the previous 99.

So either by adding 100 each time or by using the general formula for calculating the 𝑛th term of an arithmetic sequence, we’ve found that the first six terms in the sequence representing Mrs. Jones’ monthly account balances are 3250, 3350, 3450, 3550, 3650, and 3750.