WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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Today we are going to talk about geometric series.
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In the previous lesson, I introduced the concept of geometric sequences; so this continues on with that knowledge.
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So, what are geometric series? A **geometric series** is the sum of the terms in a geometric sequence.
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Again, make sure that you have geometric sequences learned, that you understand that well, before going on to geometric series.
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But just briefly, recall that a geometric sequence is a list of numbers.
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And what is unique about this list is that you find one term by multiplying the previous term by a number r, which is the common ratio.
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For example, here the common ratio is 2: so I multiply 8 times 2 to get 16, times 2 is 32, times 2 is 64.
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You can always find that common ratio by taking a term and dividing it by the previous term.
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This is a geometric sequence: today we are going to move on to talk about geometric series.
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And a geometric series is the sum of the terms; so from this, we could get a geometric series, 8 + 16 + 32 + 64.
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This is the geometric series: first term + second term + third term, and on and on, until we get to that last term.
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Now, this is a finite series, but you could also have an infinite series, where it just continues on indefinitely.
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So, what we want to find, often, is the sum of a particular number of terms in the series.
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Now, I could look up here and say, "OK, I want to find the first three terms: that is 8 + 16 + 32."
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And then, I could just add that up and figure out what it is.
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But that is going to get really cumbersome to add manually; so we have a formula for the sum of a geometric series.
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The sum of the first n terms of a geometric series is given by this formula.
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And we have the limitation that r does not equal 1, because if r equaled 1, if I had, say, 3,
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and I just multiplied it by the common ratio of 1, I would just get 3 again and again and again, and it wouldn't really change.
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So, the limitation is that the common ratio cannot equal 1.
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Looking at an example, 6 + 18 + 54 + 162, let's say I wanted to find the sum of this entire series--the sum of all the four terms.
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I could use this formula: my first term here is 6--what is my common ratio?
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Well, I can say 18/6 is 3; so the common ratio is 3.
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Therefore, the sum of these four terms would be 6, times (1 - 3^n) (n = 4 in this case),
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divided by (1 - 3); this is going to give me 6(1 - 3⁴).
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Well, recall that 3 times 3 is 9, times 3 is 27, times 3 is 81; so this is 81, divided by 1 - 3.
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So, the sum is 6 times...1 - 81 would give me -80, divided by...1 - 3 is -2.
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Let's go up here to continue on: 6 times -80 is going to give me -480, because 6 times 8 is 48; add a 0;
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divided by -2; the negatives cancel out; 480/2 is 240.
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So, I was able to find this sum by using a formula, rather than just adding each number.
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And the formula requires that I know the common ratio, the first term, and n.
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Sigma notation: as with arithmetic series, we can also use sigma notation as a concise way to express a geometric series.
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Again, the Greek letter Σ means sum; and the variable that we are going to use is called the index.
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So here, I have a lower index; and then the upper index tells me how high to go.
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So, here I have n going from 1 to 5; and then I am going to have the formula for the general term
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written right here, so I know how to find each term in this geometric series.
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Looking at an example that is specific: as I said, they often use the letter i for the index in sigma notation, so I am going to use i.
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i going from 1 to 6...and the formula to find each term is going to be 3 raised to the i power.
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Therefore, I can find this series; it is going to be 3 raised to the first power, plus 3...
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I started out with 1, and I inserted that here; next I am going to go to 2.
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Then, I am going to go to 3, then 4, 5, and 6.
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And then, you could, of course, figure this out; this is 3 + 9 + 27 + 81 + 729 + 2187.
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So, this notation means this; and again, this is just a different way of writing a geometric series.
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But the concepts that we have discussed remain the same.
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All right, we learned one sum formula: and there is a second formula.
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This second formula is very valuable when you know the first and last terms, and you know the common ratio, but you don't know the number of terms.
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If n is not known, use this formula.
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For example, maybe I have a series, and I know that the first term is 128; that the last term is 4; and that the common ratio is 1/2.
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But I don't know n--n is not known--and I am asked to find the sum.
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I can do that using this formula: the sum is going to be the first term, minus this last term, times r, divided by (1 - r).
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which is going to be equal to 128 minus...4 times 1/2 is 2...divided by...1 minus 1/2 is just 1/2.
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This is going to be equal to 126 divided by 1/2, and that is the same as 126 multiplied by the reciprocal of 1/2, which is 2.
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And that is 252; so the sum is going to be 252, and I was able to find that because I knew the first term; I knew the last term; and I knew the common ratio.
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And I had a formula that did not require me to know n.
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All right, the formula for s < font size="-6" > n < /font > can be used to find a specific term in the series.
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And a very important term is the first term: so we often use s < font size="-6" > n < /font > to find the first term in a series.
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So, looking at one of the sum formulas that we just discussed, that would be the first term, times (1 - r^n), divided by (1 - r).
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Let's say that you are given that the sum is 62.
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And you are also given that the common ratio is 2, and that the number of terms is actually 5.
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And you want to find the first term; you can do that with this formula.
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I know that 62 equals the first term, times 1 - 2, raised to the n power (here, n is 5), times 1 - 2.
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So, 62 = a₁...2 to the fifth...2 times 2 is 4, times 2 is 8, times 2 is 16, times 2 is 32; so that is 32.
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1 - 32, divided by -1...therefore, this equals...rewriting this as 62 =...the first term...1 - 32 would give me -31,
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divided by -1; if you just pull the negative out front, then those are going to end up canceling; -31a/-1 is just going to give you a positive.
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So, I am going to rewrite this as 62 = 31 times the first term, because this is just a -1 down here, and I can rewrite this much more simply.
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OK, dividing both sides by 31 gives me that the first term equals 2.
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So, I was able to find the first term by being given the sum and the common ratio and the number of terms.
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Frequently, you will use one of your sum formulas to find the first term in a geometric series.
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All right, on to Example 1: Find the sum, s < font size="-6" > n < /font > , for the geometric series with a first term of 162,
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with a common ratio of 1/3, and with a number of terms equal to 6.
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All right, the formula for the sum: a₁(1 - r^n)...and since I know n, I can use this formula,
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rather than the other one...divided by (1 - r); and I want to find s < font size="-6" > n < /font > .
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s < font size="-6" > n < /font > equals the first term, which is 162, times (1 - 1/3⁶), all of that divided by (1 - 1/3).
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Therefore, the sum equals...162 times 1, minus 162 times 1/3 to the sixth power, divided by...1 - 1/3 is going to leave me with 2/3 in the denominator.
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Now, this looks like it might be complicated to work with; but there is a shortcut.
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If you think of...I have my 162; I can think of 162 as 2 times 81, and 81 is a multiple of 3, so you might be able to see where I am going with this.
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2 times 81...as I said, that is a multiple of 3; 3 times 3 is 9, times 3 is 27, times 3 is 81.
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Therefore, I could rewrite this as 162 - 2...let's move this out of the way a bit...times 3 to the fourth power, times 1/3 to the sixth power.
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I can use rules governing how I work with powers, and say, "OK, if this gives me 3⁴/3⁶, 162 -2 times...
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if I look at this as 3⁴/3⁶, they have the same base; if I take 6 - 4, this is going to give me 1/3²."
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So, this is 2 times (1/3)², divided by 2/3--it is much easier to work with now.
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This is simply going to be 162 minus 2 times 1/9, or 2/9; all divided by 2/3.
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Coming up here to finish this out: the sum, therefore, equals 162 - 2/9; that is going to give me 161 and 7/9, all divided by 2/3.
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Recall that, then, dividing by 2/3 is the same as multiplying by 3/2.
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That is not a very pretty answer, but you can simplify this, calculate it out, use a calculator...but this does give the sum.
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So again, I was given the first term, the common ratio, and the number of terms.
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I can use this formula, since I have the number of terms.
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This looked like it was going to be very messy to work with: 162 times (1/3)⁶.
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But by recognizing that I could break 162 into 2 times a multiple of 3, I was able to get this into a base with 3, 3⁴.
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Dividing 3⁴ by 3⁶ canceled out, and I got (1/3)²;
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and that simplified things a lot to give me this answer that I have in the upper left.
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Find the sum of the first 8 terms of the geometric series.
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So, here we are asked to find the sum of the first 8 terms of this series, and that would be s₈.
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Thinking about which formula I am going to use: I know n; since I am looking for the first 8 terms, I know n.
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I can also easily find r, because I can just take 1 divided by 1/4; 1 divided by 1/4 is the same as 1 times 4, or 4.
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So, I have r; I have n; the other thing is the first term, and I have that.
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With these three pieces of information, I know that I can use the formula
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that the sum equals the first term, times (1 - r^n), divided by (1 - r).
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So, the first 8 terms...that is going to be 1/16, times 1 minus...r is 4...raised to the n power, where n is 8, divided by (1 - 8).
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OK, let's look at what I have: I have 1/16 times 1, minus 1/16 times 4⁸.
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Actually, this is not 1 - n; it is 1 - r; so let's go ahead and change that to a 4.
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All right, and then I have 1/16 times 4⁸; as with the previous problem,
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you might recognize that there is something to make this a lot easier than taking 4 to the eighth power and dividing it by 16.
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You are going to recognize that you can rewrite this 1/16 as (1/4)².
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So, 1/4 times 1/4 would give me 1/6; that times 4⁸ allows me to do some canceling to make things simpler.
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1 - 4 just gives me -3; let's go up here and work this part out.
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This is (1/4)² times 4⁸; this is the same as taking 4⁸ and dividing it by 4².
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Using my rules governing exponents, 4⁸/4²...if I want to divide, and I have like bases,
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I just subtract the exponents; so this is going to give me 4 raised to the sixth power.
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So, knowing your rules of exponents is important to solve when you are working with this type of problem.
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Therefore, I am going to get 1/16 - (1/4)² times 4⁸
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(that is the same as 4⁶, so I am going to go ahead and rewrite it that way) divided by -3.
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At this point, you are going to have to do a lot of multiplying, or else use your calculator to determine that 4⁶ is actually 4096.
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And we are dividing by -3; so the sum is 1/16 - 4096; that comes out to -4095 and 15/16, all divided by -3.
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And a negative and a negative will give me a positive, so that would give me 4095 and 15/16, divided by 3.
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You can work this out with your calculator and see that this sum is approximately equal to 1365.3.
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This does give us our answer; but we can estimate to put it in a neater decimal form.
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Finding the sum: this time, we are given the first term; we are given the last term; and we are given the common ratio.
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But we don't know the number of terms--we don't know what n is.
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But that is OK, because we have that other formula that allows us to find the sum of a geometric series when we know the first term;
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we know the last term; and we know the common ratio (that second formula that we worked with).
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All right, looking for the sum: I have the first term; this is 4 - a < font size="-6" > n < /font > (which is 8748), times the common ratio of 3, all of that divided by 1 - 3.
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That is going to give me that the sum equals 4 minus...multiply this out, or use your calculator, to get 26244, divided by 1 - 3 (gives me -2).
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Coming up here to the second column: the sum is going to be equal to 4 - 26244, or -26240, divided by -2.
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A negative divided by a negative gives me a positive; and 26240 is even--I divide it by 2, and I get a nice whole number, 13120.
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So, the sum of this geometric series is 13120; and I found that using the formula
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that only requires that I know the first term and the last term, and the common ratio.
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I didn't have to know the number of terms in the series, and I could still find the sum of the terms.
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Example 4: We are asked to find the first term.
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As I mentioned, it is important to have the first term frequently when you are working with these series.
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And you can use the sum formulas to find that first term.
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I am going to look at the information I have: I have the sum; I have the common ratio; and I have the number; but I don't have the last term.
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So, I use the formula that involves the number of terms, not the last term.
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And what I am looking for is the first term; therefore, the sum is 1020, equals the first term,
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times 1 - r (is 2), and it is 2 raised to the eighth power, divided by 1 - 2.
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And it is 2 raised to the eighth power, divided by (1 - 2); this is going to give me 1020 equals the first term, times 1 minus...
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if you go through your powers of 2, you will find that 2 times 2 is 4, times 2 is 8, times 2 is 16, times 2 is 32.
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So, 2⁵ - 32; then, we are going to get 2⁶ is 64; 2⁷ is 128; and then, 2⁸ is 256.
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That is 1 - 256, divided by 1 - 2 (is -1); therefore, 1020 = a₁...1 - 256 is -255; divided by -1.
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Therefore, 1020 = -a₁ times 255, divided by -1.
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Well, a negative and a negative gives me a positive, so that is 1020 = a₁ times 255.
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I just take 1020 and divide by 255; so I have divided both sides by 255.
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And I am going to get that the first term equals 4.
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I was asked to find the first term, and I found that using the sum formula requiring the first term, the common ratio, and the number of terms.
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And I determined that the first term in this geometric series is 4.
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That finishes up this lesson on geometric series on Educator.com; thank you for visiting!