Hint
Relate \(\displaystyle S_{2}\) to S_∞ through \(\displaystyle 1-r^{2}\). The requested error is the tail, not the last retained term.
Worked solution
(a)
\(\displaystyle \frac{a}{1- r}= 24\) and \(a(1+r)=18\). Therefore \(24(1-r^2)=18\), giving \(\displaystyle r^{2}= \frac{1}{4}\). The stated positive ratio selects \(\displaystyle r= \frac{1}{2}\); then \(\displaystyle a=24(1-r)=12\).
Answer: \(\displaystyle a= 12, \ r= \frac{1}{2}\).
(b)
The finite sum is \(\displaystyle \frac{12(1- (\frac{1}{2})^{n})}{\frac{1}{2}}= 24(1- 2^{- n})\). Subtracting it from 24 leaves \(R_n=24\cdot2^{-n}\).
Answer: \(S_n=24(1-2^{-n}),\ R_n=24\cdot2^{-n}\).
(c)
\(\displaystyle 24\cdot 2^{−n}<0.1\) means \(\displaystyle 2^{n}>240\). The powers \(\displaystyle 2^{7}=128\) and \(\displaystyle 2^{8}=256\) straddle this bound. \(\displaystyle R_{7}=0.1875\) and \(\displaystyle R_{8}=0.09375\). Since \(\displaystyle R_{n}\) decreases, 8 is the first successful index.
Answer: \(n=8\).
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.
(a) · 3 marks
- 1 mark: Relate the two sums using correct GP formulas.
- 1 mark: Obtain \(\displaystyle r= \frac{1}{2}\), choosing the stated positive root.
- 1 mark: Obtain \(\displaystyle a=12\).
Alternative methods: Elimination from a+ar=18 and \(\displaystyle a=24(1-r)\) is accepted.
Common error: Keeping \(\displaystyle r= - \frac{1}{2}\) despite 0<r<1.
(b) · 2 marks
- 1 mark: Give the correct expression for \(\displaystyle S_{n}\).
- 1 mark: Give the correct expression for \(\displaystyle R_{n}\).
Alternative methods: Writing the tail as the first omitted term divided by 1−r is accepted.
Common error: Using the nth term itself for the whole omitted tail.
(c) · 3 marks
- 1 mark: Obtain \(\displaystyle 2^{n}\)>240 or an equivalent inequality.
- 1 mark: Give \(\displaystyle n=8\).
- 1 mark: Check the strict bound and minimality using consecutive values/powers.
Alternative methods: A logarithm calculation with a valid strict integer check is accepted.
Common error: Reversing the inequality or accepting a remainder equal to the tolerance.
Total: 8 marks.