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OCR A level C2 maths, Jan08 Q8

Last part is proper dodgy :eek3:

8) (iii) The sum of the first N terms is denoted by S(to base N), and the sum to infinity is denoted by S(to base sideways 8:p: )

Show that the inequality S(base sideways 8) - S(base N)<0.01 can be written as:
0.8^N<0.0002
and use logaritms to find the smallest possible value of N

this is 7 marks and step by step on how to do it wud be useful :woo:
Reply 1
Avatar for ttoby
ttoby
15
Could you post the whole question including (i) and (ii) since it will contain details needed to answer this part.
Reply 2
Avatar for RTJohnson92
RTJohnson92
OP
2
yep right...

8) The first term of a geometric progression is 10 and the common ratio is 0.8

(i) Find the fourth term

(ii) Find the sum of the first 20 terms, giving your answers correct to 3s.f

(iii)The sum of the first N terms is denoted by S(to base N), and the sum to infinity is denoted by S(to base sideways 8 )

Show that the inequality S(base sideways 8) - S(base N)<0.01 can be written as:
0.8^N<0.0002
and use logaritms to find the smallest possible value of N
Reply 3
Avatar for ttoby
ttoby
15
S∞−SN<0.01S_\infty -S_N <0.01
Substituting in the formulas for sum to infinity and sum to N,
101−0.8−10(1−0.8N)1−0.8<0.01\dfrac{10}{1-0.8} - \dfrac{10(1-0.8^N)}{1-0.8}<0.01
Taking out a factor of 10/(1-0.8)=10/0.2,
100.2(1−1+0.8N)<0.01\dfrac{10}{0.2} (1-1+0.8^N) <0.01

Now you can finish the question by rearranging this to make 0.8^N the subject.

For the second part, take logs of both sides and rearrange the left hand side to look like N log 0.8. Then divide both sides by log 0.8 (you'll need to change the sign of the inequality because log 0.8 is negative - check it on your calculator if you're not sure).

Type in the right hand side into your calculator and this will tell you that N is greater than some decimal number. Since N has to be an integer, find the smallest integer that's just bigger the the decimal you found.
Reply 4
Avatar for RTJohnson92
RTJohnson92
OP
2
ttoby
S∞−SN<0.01S_\infty -S_N <0.01
Substituting in the formulas for sum to infinity and sum to N,
101−0.8−10(1−0.8N)1−0.8<0.01\dfrac{10}{1-0.8} - \dfrac{10(1-0.8^N)}{1-0.8}<0.01
Taking out a factor of 10/(1-0.8)=10/0.2,
100.2(1−1+0.8N)<0.01\dfrac{10}{0.2} (1-1+0.8^N) <0.01

Now you can finish the question by rearranging this to make 0.8^N the subject.

For the second part, take logs of both sides and rearrange the left hand side to look like N log 0.8. Then divide both sides by log 0.8 (you'll need to change the sign of the inequality because log 0.8 is negative - check it on your calculator if you're not sure).

Type in the right hand side into your calculator and this will tell you that N is greater than some decimal number. Since N has to be an integer, find the smallest integer that's just bigger the the decimal you found.


thanks for all of this mate!!

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