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    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?
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    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
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    (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
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    (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.
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    (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.
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    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?
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    (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?
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    (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!
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    (Original post by Lucofthewoods)
    Sure!
    Yours or mine baby?
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    (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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