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# Worked example: arithmetic series (recursive formula)

Sal evaluates the sum of the first 650 terms in the sequence defined recursively as {aᵢ=aᵢ₋₁+11, a₁=4}. He does that by finding the 650th term and using the arithmetic series formula (a₁+aₙ)*n/2.

## Want to join the conversation?

• What is the difference between a series and a sequence?
• In a series, the individual terms are being added to each other.
A sequence just lists each individual term and uses commas to separate them.
• Isn't it easier (in evaluating this) to simply convert the recursive expression to an explicit one? Then you can just plug the values you want in, and solve for the answer. (With a bit of algebraic manipulation, of course.)
• i jus need to know how to write the equation
• When finding n (number in a series) when x is the rate of change, is this a correct formula:
n=|(an-a1)/x|+1
• You are correct but, in this example, solving for n was not necessary since we were provided with it at the beginning. Sal told us that there are 650 terms.
• why does sal add 4 twice into the calculation where the formula only asks for it once.
• they first time, he added 4 as part of the formua for a(n) ie the 650th term
he second time it was as part of the formula for S(n) ie sum of 650 terms
look through the video once again im sure u'll get it
hope it helps OwO
• When Sal writes a shouldn't it be, if a = 4, 3+11
I-1
• a1 or the first term is 4 and the difference between consecutive numbers is 11.
• I don't understand how we got 649, can someone explain?
• Ok, so think about it this way. I'm adding 650 terms in a sequence, right? Ok, so I'm given the first term, right? Yes, I am. If I'm given the first term, then I don't do anything to it, because it's already defined. This means that I have to change all but the first term, which is given. So, you take that first one from 650. 650-1, obviously, is 649, and that's how he got that number. Hope this helps.
• If the last number of ai= is minus, do you add 1 to the a650?
• NO you minus. Hope this helps
(1 vote)
• Can we find the 650th term by using an= a + (n-1)d, where a is the first term and d is the common difference?
• A simple way to generate a sequence is to start with a number a, and add to it a fixed
constant d, over and over again. This type of sequence is called an arithmetic sequence.
Definition: An arithmetic sequence is a sequence of the form
a, a + d, a + 2d, a + 3d, a + 4d, …
The number a is the first term, and d is the common difference of the
sequence. The nth term of an arithmetic sequence is given by
an = a + (n – 1)d
The number d is called the common difference because any two consecutive terms of an
arithmetic sequence differ by d, and it is found by subtracting any pair of terms an and
an+1. That is
d = an+1 – an
Is the Sequence Arithmetic?
Example 1: Determine whether or not the sequence is arithmetic. If it is arithmetic, find
the common difference.
(a) 2, 5, 8, 11, …
(b) 1, 2, 3, 5, 8, …
Solution (a): In order for a sequence to be arithmetic, the differences between
each pair of adjacent terms should be the same. If the differences
are all the same, then d, the common difference, is that value.
Step 1: First, calculate the difference between each pair of adjacent
terms.
5 – 2 = 3
8 – 5 = 3
11 – 8 = 3
Step 2: Now, compare the differences. Since each pair of adjacent terms
has the same difference 3, the sequence is arithmetic and the
common difference d = 3.
(1 vote)
• Whats the use of recursive formulas?
(1 vote)
• It's just another way to describe things, and one that can come up very naturally.