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Worked examples: finite geometric series

Get a front row seat to Sal's step-by-step solutions to finite geometric series. Learn how to find the sum of the first 50 terms of a series by multiplying each term by a common ratio. Discover how to apply the formula for the sum of a finite geometric series. Plus, explore how to handle series with a change in sign.

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• this thing is sooo confusing
• I know this is an old post, but for the newbies who are still confused this is what I've understood so far:

Basically for the first few minutes he used the formula described in the previous video "Finite geometric series formula". That formula was plugged in for the first few minutes.
For example,
In the first problem he uses the formula and then to simplify it further he simplifies the denominator which should be simple enough (1-10/11 = 1/11) and then he multiplies the entire thing by 11 to get rid of the denominator. After doing this he's simplified it enough to get the SUM OF ALL OF THE TERMS. That's what the formula is for.
If you can understand this much, it's easier to carry on with the other problems. If you don't understand, go to the previous videos in the Algebra 2 course - Unit 3 - Lesson 7 (Videos 1 and 2). That's your best bet to understand it.
• if you keep on multiplying 1 by 10/11 for an infinite amount of times, would it eventually become zero? Thanks in advance!
• That is the correct intuition to have. Obviously, we don't reach 0 after any finite number of multiplications, but we say the product tends to 0 or approaches 0.
• for the last problem wouldnt it be to the 31rst power?
• Good question. It's a little complicated, but:
The formula for the sum was made for a sequence of n terms. If there are 30 terms (n=30), and if the 1st term is 10, you multiply by 9/10 from i=2 to i=30 (29 terms - so you have (9/10)^29) plus 1 more for i=1 (the first term, 10) makes 30.
So if he had set it up like the earlier problem "10 + 10(9/10) + 10(9/10)^2 + ... + 10*(9/10)^29" then you add 1 to the 29. But when he just says "there's 30 terms" or "29 terms", you don't add anything.
• At , it says a sub i is equal to a sub i minus i times 9/10. Is this supposed to be i-1?
• It's confusing... the dot is actually the multiplication symbol, not a dot on a "i".
So, Sal does have "i-1"
• this is so confusing 😭
• Can someone help me understand how to find the first term when given the common ration and sum of the first n terms? I'm having trouble with the solving part. I get to the equation, but then it gets complicated trying to find a with fractions and exponents in the equation. Thanks in advance!
• a=S(1-r)/(1-r^n).(S-sum till n terms, n-number of terms, r-common ratio,a-first term)
If you have to find the first term given the sum, and common ratio, substitute the value of S, r and n and do the calculations carefully. You will get the value of a with this method.
• What is the difference between finite/infinite and convergent/divergent geometric series?
• A finite geometric series is a series of the form sum n=0 to k of ar^n. Note that this type of series has finitely many terms, and so the sum always exists.

An infinite geometric series is a series of the form sum n=0 to infinity of ar^n. Note that this type of series has infinitely many terms. For this type of series, the question of convergence vs. divergence arises.
Note that the sum of this type of series is defined as the limit of the sequence of partial sums (that is, sums of finite series of the form sum n=0 to k of ar^n) as the number of terms approaches infinity. If this limit of partial sums exists (which occurs when either a = 0 or |r| < 1), then the infinite geometric series is convergent and the sum equals the value of this limit. If this limit of partial sums does not exist (which occurs when both a is nonzero and |r| >= 1), then the infinite geometric series is divergent and the sum does not exist.

In summary: finite/infinite refers to the number of terms of the series, but convergent/divergent refers to the type of limiting behavior of the sequence of partial (finite) sums for an infinite series.

Have a blessed, wonderful day!
• When Sal simplifies 1(1- -.99^80) why does he simplify this portion to 1-.99^80 instead of 1+.99^80? It's written at
• You may have forgotten to look at the parenthesis. As Sal writes it, it is (1 - (-0.99)^80). Since 80 is an even exponent, both -0.99 and +0.99 will output to the same number, so you can effectively cancel out the second negative sign, giving you 1 - 0.99^8. Since exponentiation happens before addition, we can do this and canceling out the two negative signs would be wrong.
• Wait at , I thought dividing by 1/10 was multiplying by 10 not 100??