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Course: Class 10 (Old) > Unit 5
Lesson 1: Intro to arithmetic progressionsUsing arithmetic sequences formulas
Google ClassroomSal finds terms of arithmetic sequences using their explicit and recursive formulas.
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The video sort of makes sense, but when I try the problems I get wildly wrong answers. How does one make sense of this.(20 votes)
It is hard to say what is going wrong without some specific example(s). So if you can give a problem and how you are solving it and getting a wildly wrong answer, people can analyze this and help explain where to go from there.(15 votes)
why are there random numbers whyyy please explain what is happeninnng(5 votes)
The second question can also be done as follows:
Given:
a1 = -7
ai = a(i-1) -2
=> common difference = d = -2
We know that:
an = a1 + (n-1)d
=> a5 = -7 + (4)-2 = -7 - 8 = -15(5 votes)
Is there a short cut to this? Let's say we had the same equations, but instead of looking for the 5th term, we were to look for the 16th term? It would take a really long time to get there, and this method is pretty slow.(4 votes)
Yes... if you keep working thru the lessons on sequences, you will see techniques for calculating specific higher terms in the sequence.(4 votes)
If you're given two successive terms, say 13 and 18, how do you find the 50th term? Can I make an equation like is given in the first problem?(2 votes)
There isn't a way to find the 50th term if you are only given two terms. You need at least 3 terms to be able to know what the pattern is.
In another answer, they say that you can just add 5 to the previous term to find the next term. However, it's just as valid to say you can multiply by 18/13 to get the next term. Or you can say you add 5*(the placement of the number) to get the next number (which would lead to 13, 18, 28, 43). While "adding 5 to the previous number" is probably the pattern, it doesn't have to be, and you shouldn't assume that it is on a test... or in real life.(5 votes)
I understand the Recursive Formula, but is there a faster way do do it??(2 votes)- The faster way is to convert the recursive formula into an explicit formula which lets you get any term with minimal effort.(4 votes)
I got the problem a(n)=-6-4(n-1) Find the fourth term in sequence.
I got (-)30, but it said the answer is (-)18.
Heres my reasoning:
(4-1)=3
-6-4 =-10
-10*3=-30
Please help me. Thank you!(2 votes)
You didn’t follow the order of operations. So what you did was (-6-4)*3, but what you need to do is -6-4*3. So you multiply 4*3 first to get 12, then take -6-12=-18. If you forgot the order of operations, remember PEMDAS: Parentheses, Exponents, Multiplication and Division, Addition and Subtraction.(4 votes)
This video confuses me. How can "4 + 3(i - 1)" possibly define an arithmetic sequence? It relies on multiplication, not addition, and therefore there is no common difference between terms.(1 vote)- Remember, multiplication is repetitive addition. As you go term to term, you are adding the same number every time.
3+3 is the same as 3*2
3+3+3+3+3 is the same as 3*5
The common difference is the 3 and the (i-1) tells you how many 3's are getting added together.(6 votes)
Is it possible that I have this formula written as a(sub)i = -7 - 2(i-1)?(3 votes)- Where does the -1 come from? We are adding in order to be in a sequence.(2 votes)
- The -1 comes from trying to get the number before. i.e in the recursive formula, you are adding/subtracting x amount from the term before. So, to get to term 20, you add something to term 19. This is why n-1 is needed. This is also why you need to know the first term in the sequence for the recursive formula.(3 votes)
Video transcript
- [Voiceover] All right, we're told that the
arithmetic sequence a-sub-i is defined by the formula where
the ith term in the sequence is going to be 4 plus 3 times i minus 1. What is a-sub-20? And so a-sub-20 is the
20th term in the sequence and I encourage you to pause the video and figure out what is the 20th term? Well, we can just think
about it like this. A-sub-20, we just use this
definition of the ith term. Everywhere we see an i,
we would put a 20 in, so it's going to be 4 plus 3 times 20 minus 1, so once again, just to be clear, a-sub-20, where instead of a-sub-i,
wherever we saw an i, we put a 20, and now we can just compute
what this is going to be equal to. This is going to be equal to 4 plus 3 times 20 minus 1 is 19. 3 times 19, let's see, 3 times 19 is 57, right? It's 30 plus 27, yep. This is 57, and 4 plus 57 is equal to 61, so the 20th term in
this arithmetic sequence is going to be 61. Let's do another one of these. And here, they've told us
the arithmetic sequence a-sub-i is defined by the formula a-sub-1, they gave us the first term, and they say, every other term, so a-sub-i, they're defining
it in terms of the previous terms, so a-sub-i is going
to be a-sub-i minus 1 minus 2, so this is actually a
recursive definition of our arithmetic sequence. Let's see what we can make of this, so a-sub-5, a-sub-5 is going to be equal to, we'll use this second line right here, a-sub-5 is going to be equal to a-sub-4 minus 2. Well, we don't know what
a-sub-4 is just yet, so let's try to figure that out. So we could say that a-sub-4 is equal to, well if we use the second line again, it's going to be a-sub-3, minus 2. We still don't know what a-sub-3 is. I'll keep switching colors
'cause it looks nice. A-sub-3 is going to be equal to a-sub-3 minus 1, so a-sub-2 minus 2. We still don't know what a-sub-2 is, and so, we could write, a-sub-2 is equal to a-sub-2 minus 1, so that's a-sub-1 minus 2. Now, luckily we know what a-sub-1 is. A-sub-1 is negative 7. A-sub-1 is negative 7, so if this is negative 7, then a-sub-2 is negative 7 minus 2, which is equal to negative 9. Well that starts helping us out because if a-sub-2 is negative 9, if this is negative 9, then a-sub-3 is negative 9 minus 2, which is equal to negative 11. Well now that we know that
a-sub-3 is negative 11, so this is negative 11, we could figure out a-sub-4 is negative 11 minus 2, which is equal to negative 13. And we're almost there. We know a-sub-4, the fourth term in this
arithmetic sequence is negative 13, so, we can now, so if this is negative 13, a-sub-5 is going to be a-sub-4, which is negative 13 minus 2, which is equal to negative 15, so the fifth term in the sequence is negative 15. And we're all done.