Question

1. Find the 40th term of the sequence 2, -12, 72, ...
2. What is the sum of the first 30 terms of the series 9, 3, 1, ...?

Answer 1

40th term? That's a REALLY big number...

Answer 2

You need to know the formula for geometric sequence with this one. Do you know it?

Answer 3

It is\[a _{n}=a _{1}(r)^{n-1}\]where n is the term number you are looking for (here it's 40), a1 is the first number in the series (here it's 2), and r is the common ratio (here it's what you are doing to each number to get to the next one. Here we are multiplying each term by -6. Check it to make sure you understand.). Then fill in what you know!

Answer 4

Can you do this or do you need assistance?

Answer 5

DDCamp is right...it's a HUGE number.

Answer 6

Actually, no I take that back, it's incredibly SMALL!!!! Cuz it's negative!

Answer 7

is it -4.456^30???

Answer 8

Yes, it is! did you know that already or did the formula help? Jw.

Answer 9

What about the next one, the sum of the geometric series?

Answer 10

The formula really helped! Could you possibly help me with the second one? I'm really stuck on that one.

Answer 11

So i got the sequence is dividing by 3 each term, and now I'm just really confused.

Answer 12

Looking right now, ok?\[S _{n}=\frac{ a _{1}(1-r ^{n}) }{ 1-r }\]The same thing applies here but with a different formula. n is the number of terms (here it's 30), a1 is the first term in the series (here it's 9), and r is the ratio or what do you do to the second number to get it to the first (just like what do you do to the third to get it to the second). Here this one is a bit tricky, because in order to get from a 3 to a 9, you have to divide by something. That's the tricky part. You have to divide 3 by 1/3. When you do that, you get 9. So r here is 1/3. Fill in your equation now and see what comes out! And hope it's right the way I told you to dot it! lol

Answer 13

So would it be 536870912/7625597484987? or 0.000007.....???

Answer 14

It would be very large becasue you are dividing by 1/3 which is actually multiplying by 3 when you flip the fraction and change the division sign to a multiplication sign. Its huge.

Answer 15

Ok, so Im stuck again! haha I'm not sure what I did wrong! lol

Answer 16

Now I'm stuck, too. Lol. Seriously! Maybe I'm misremembering my formula! Let me look into it. Just a sec.

Answer 17

so.... what's the "common ratio" or the multiplier from term to term?

Answer 18

divide by 3?

Answer 19

hmm is supposed to be a multiplier ...... no quite is not 1/3 if that's what you meant

Answer 20

OMG I'm so dumb! I'm trying to find out why my number seems so strange but I am supposed to be looking for the sum of the terms and instead I was trying to find out the 30th term. Ok, duh, here's what you do. \[S _{n}=\frac{ a _{1}(1-r ^{2}) }{ 1-r }\]Filling in our info we get\[S _{n}=\frac{ 9(1-\frac{ 1 }{ 3 }^{2}) }{ 1-\frac{ 1 }{ 3 } }\]When you take 1/3 to the 30th power you get .000000000000004586. Seriously. So when you subtract that from 1 you will get 1. So then you multiply that 1 by the 9 to get\[S _{n}=\frac{ 9 }{ 1-\frac{ 1 }{ 3 } }\]Do the math and get 27/2, which is 13.5

Answer 21

Ahh thank you so much!!!

Answer 22

You're very welcome!