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p1-Sequences+series watch

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    The sum of the first 3 terms of a GP is 65/9 and the sum to infinity is 7.5.
    Next the qu asked me to find out a and r which i did...r=1/3, a=5.
    Then it says find the least number of terms required so that their sum differs from 7.5 by less than 10^-4....heeeeeyyyylllppp! trial ad error meth?
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    (Original post by BossLady)
    The sum of the first 3 terms of a GP is 65/9 and the sum to infinity is 7.5.
    Next the qu asked me to find out a and r which i did...r=1/3, a=5.
    Then it says find the least number of terms required so that their sum differs from 7.5 by less than 10^-4....heeeeeyyyylllppp! trial ad error meth?
    haven't looked at this question yet, but just to let you know, I answered the other sequences and series question. It's probably on the second page by now.
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    (Original post by mockel)
    haven't looked at this question yet, but just to let you know, I answered the other sequences and series question. It's probably on the second page by now.
    oohhh thanks very much mockel, i forget about that one! ill go check
    oh and rep coming your way if its right!
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    anyone got ideas for this 1?
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    (Original post by BossLady)
    Then it says find the least number of terms required so that their sum differs from 7.5 by less than 10^-4.
    Which sum differs from which?
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    Sn = 7.5[1-3^(-n)] you're looking for 7.5 - Sn < 10^-4 so 7.5(1-[1-3^(-n)]) < 10^-4, 7.5*3^-n < 10^-4 rearrange to 3^n > 7.5*10^4, take logs of both sides n log3 > 4 + log7.5, n > (4 + log7.5)/log3 = 10.21... so you need to sum up to and including n=11, ie 11 terms
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    (Original post by Bezza)
    Sn = 7.5[1-3^(-n)]
    where do u get this from,,,,, i thought the formula was
    s=a(1-r^n)/(1-r^n)
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    (Original post by BossLady)
    where do u get this from,,,,, i thought the formula was
    s=a(1-r^n)/(1-r^n)
    No - if it was the 1-r^n would cancel Sn = a(1-r^n)/(1-r) = 5[1-3^(-n)]/(1-1/3) = 5[1-3^(-n)]/(2/3) = 15/2*[1-3^(-n)] = 7.5[1-3^(-n)]
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    (Original post by Bezza)
    No - if it was the 1-r^n would cancel Sn = a(1-r^n)/(1-r) = 5[1-3^(-n)]/(1-1/3) = 5[1-3^(-n)]/(2/3) = 15/2*[1-3^(-n)] = 7.5[1-3^(-n)]
    lol sorry i meant [1-r] on the bottom of frac!! oh and thanks bezza
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    (Original post by BossLady)
    lol sorry i meant [1-r] on the bottom of frac!! oh and thanks bezza
    Yeah, yeah sure you did
 
 
 
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