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# p1-Sequences+series watch

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1. 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?
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.
3. (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!
4. anyone got ideas for this 1?
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?
6. 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
7. (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)
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)]
9. (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
lol sorry i meant [1-r] on the bottom of frac!! oh and thanks bezza
Yeah, yeah sure you did

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