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# Geometric progressions watch

1. Sorry- I'm awful at these

Find the sum of the first 10 terms of a GP which has 3rd term 20 and 8th term 640.

so I've worked out that ar^2 is 20 and ar^7 is 640
So what now?
2. thats a good start, you can divide the terms and find the value of r. then get the value of a. then use the sum of a GP formula
3. (Original post by jitinder)
thats a good start, you can divide the terms and find the value of r. then get the value of a. then use the sum of a GP formula
How do you divide with a and r though since there are 2 unknowns
4. (Original post by Lucofthewoods)
Sorry- I'm awful at these

Find the sum of the first 10 terms of a GP which has 3rd term 20 and 8th term 640.

so I've worked out that ar^2 is 20 and ar^7 is 640
So what now?
(Original post by jitinder)
thats a good start, you can divide the terms and find the value of r. then get the value of a. then use the sum of a GP formula
Elaboration:
Since we know a cannot be zero (GP so you're multiplying by r, if a was 0 then you'd keep getting a) Similarly r cannot be zero.
This means we can divide by a and/or r.
Treat these now as simultaneous equations. Hint: you want to cancel the a and a few r's.
5. (ar^7)/(ar^2) = 640/20.

You'll see that "a" will be eliminated giving you r^5 = 32. meaning r = 2.

then substitute the value of r in any one and get a. etc.
6. If F(x) is a solution of this functional equation for a given f(x) in a geometric progression, then so is F(x)+C(x) for any periodic function C(x) with period 1. Therefore each indefinite series actually represents a family of functions. However the solution equal to its Newton series expansion is unique up to an additive constant C.

Hope this helps
Btw will you sleep with me?
7. (Original post by jitinder)
(ar^7)/(ar^2) = 640/20.

You'll see that "a" will be eliminated giving you r^5 = 32. meaning r = 2.

then substitute the value of r in any one and get a. etc.
Sheesh... Let him figure it out?
8. (Original post by JackScholey)
If F(x) is a solution of this functional equation for a given f(x) in a geometric progression, then so is F(x)+C(x) for any periodic function C(x) with period 1. Therefore each indefinite series actually represents a family of functions. However the solution equal to its Newton series expansion is unique up to an additive constant C.

Hope this helps
Btw will you sleep with me?
Sure!
9. (Original post by Lucofthewoods)
Sure!
Yours or mine baby?
10. (Original post by carpetguy)
Sheesh... Let him figure it out?
Oops :P But hey, there is still some brainwork left :P

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