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Help with Geometric Series question

Hey guys. The question is:

"The third and fifth terms of a geometric sequence are 32 and 512. Find the first term and the common ratio."

I have tried doing 512/32 and then dividing by 2 to get the fourth term but it doesn't work. :frown:

Also, I have another question that I have done but I'm not too sure if I've got the correct answer:

"Find the first three terms of a geometric sequence whose sum to n terms is given by 5/2(3^n-1)."

I got my answer as 5, 15, 45. It would be great if one of you guys checked it.

Thanks. :smile:
Reply 1
Avatar for Zacken
Zacken
22
Original post by SuperSNESLiam
Hey guys. The question is:

"The third and fifth terms of a geometric sequence are 32 and 512. Find the first term and the common ratio."

I have tried doing 512/32 and then dividing by 2 to get the fourth term but it doesn't work. :frown:

Also, I have another question that I have done but I'm not too sure if I've got the correct answer:

"Find the first three terms of a geometric sequence whose sum to n terms is given by 5/2(3^n-1)."

I got my answer as 5, 15, 45. It would be great if one of you guys checked it.

Thanks. :smile:


Why would you divide by two? Let the first term be a and the common ratio r. Then the third term is ar^2 and the fifth term is ar^4.

So the ratio is ar^4 / ar^2 = r^2 = 512/32

Now you square root, not divide by two...
Original post by Zacken
Why would you divide by two? Let the first term be a and the common ratio r. Then the third term is ar^2 and the fifth term is ar^4.

So the ratio is ar^4 / ar^2 = r^2 = 512/32

Now you square root, not divide by two...


Thank you.

As to why I divided by two: ¯\_(ツ)_/¯
Original post by Zacken
Why would you divide by two? Let the first term be a and the common ratio r. Then the third term is ar^2 and the fifth term is ar^4.

So the ratio is ar^4 / ar^2 = r^2 = 512/32

Now you square root, not divide by two...


What about the other question?
Reply 4
Avatar for KaylaB
KaylaB
20
Original post by SuperSNESLiam
Hey guys. The question is:

"The third and fifth terms of a geometric sequence are 32 and 512. Find the first term and the common ratio."

I have tried doing 512/32 and then dividing by 2 to get the fourth term but it doesn't work. :frown:

Also, I have another question that I have done but I'm not too sure if I've got the correct answer:

"Find the first three terms of a geometric sequence whose sum to n terms is given by 5/2(3^n-1)."

I got my answer as 5, 15, 45. It would be great if one of you guys checked it.

Thanks. :smile:


So know you a3=32 a_{3} = 32 and a5=512 a_{5} = 512
And you know that the general formula for finding a term of a geometric series is an=a1rn−1 a_{n} = a_{1}r^{n-1}
So a3=a1r2=32[br]a5=a1r4=512 a_{3} = a_{1}r^{2} = 32[br]a_{5} = a_{1}r^{4} = 512
From this we can see that to get from a3 to a5 you multiply by r2
51232=16=r2⇒r=4[br]a3=32=a1r2=a142[br]32=16a1[br]2=a1 \frac{512} {32} = 16 = r^2 \Rightarrow r = 4[br]a_{3} = 32 = a_{1}r^2 = a_{1}4^2[br]32 = 16a_{1}[br]2 = a_{1}
Original post by SuperSNESLiam
Hey guys. The question is:

"The third and fifth terms of a geometric sequence are 32 and 512. Find the first term and the common ratio."

I have tried doing 512/32 and then dividing by 2 to get the fourth term but it doesn't work. :frown:

Also, I have another question that I have done but I'm not too sure if I've got the correct answer:

"Find the first three terms of a geometric sequence whose sum to n terms is given by 5/2(3^n-1)."

I got my answer as 5, 15, 45. It would be great if one of you guys checked it.

Thanks. :smile:


second part is 5,15,45. You are correct :biggrin:
Original post by DylanJ42
second part is 5,15,45. You are correct :biggrin:


Thanks.

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