This TSA checks sequences and series, AP terms and sums, GP terms and sums, recognition of AP and GP, convergence, sum to infinity, mixed AP/GP modelling, and arithmetic and geometric means.
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Sequences, Series, AP and GP Basics
1. Which statement is correct?
Tested fundamental: Difference between a sequence and a series, and basic AP/GP definitions.
Explanation: A sequence is an ordered list of terms. A series is the sum of the terms in a sequence.
Common wrong choice: C or D, from mixing up the defining properties of AP and GP.
Arithmetic Progression: Terms, Sums and Recognition
2. The first term of an AP is \(7\), and the common difference is \(4\). Which expression gives the \(n\)th term?
Tested fundamental: General term of an AP.
Explanation: For an AP, \(u_n=a+(n-1)d\). Hence \(u_n=7+4(n-1)=4n+3\).
Common wrong choice: A comes from using \(a+nd\) instead of \(a+(n-1)d\).
3. An AP is given by \[ 12,\ 9,\ 6,\ldots \] Find the \(15\)th term.
Answer format: Give your answer as an integer.
Answer: \(-30\)
Tested fundamental: Finding a term of an AP.
Explanation: Here \(a=12\) and \(d=-3\). So \(u_{15}=12+14(-3)=-30\).
Common error: Using \(15d\) instead of \((15-1)d\).
4. An AP has first term \(6\), last term \(42\), and \(10\) terms. Find the sum of the \(10\) terms.
Answer format: Give your answer as an integer.
Answer: \(240\)
Tested fundamental: Sum of an AP using the first and last terms.
Explanation: Use \(S_n=\frac{n}{2}(a+l)\). Hence \(S_{10}=\frac{10}{2}(6+42)=240\).
Common error: Forgetting to multiply by \(\frac{n}{2}\).
5. An AP has first term \(5\) and common difference \(3\). Given that \(S_n=124\), find \(n\).
Answer format: Give your answer as an integer.
Answer: \(8\)
Tested fundamental: Using the AP sum formula to solve for the number of terms.
Explanation: \(S_n=\frac{n}{2}[2(5)+(n-1)3]\). When \(n=8\), \(S_8=\frac{8}{2}(10+21)=124\).
Common error: Solving for \(n\) but not checking that \(n\) must be a positive integer.
6. True or false: If the sum of the first \(n\) terms of a sequence is \(S_n\), then the \(n\)th term is always given by \[ u_n=S_n-S_{n-1}. \]
Tested fundamental: Finding \(u_n\) from \(S_n\).
Explanation: \(S_n\) contains the first \(n\) terms, while \(S_{n-1}\) contains the first \(n-1\) terms. Their difference leaves only \(u_n\).
Common wrong choice: B, from confusing \(S_n\) with \(u_n\).
7. Given \[ S_n=2n^2+3n, \] which is the common difference of the sequence?
Tested fundamental: Using \(u_n=S_n-S_{n-1}\) and identifying an AP.
Explanation: \(u_n=S_n-S_{n-1}=2n^2+3n-[2(n-1)^2+3(n-1)]=4n+1\). Since \(u_n\) is linear in \(n\), the sequence is an AP with common difference \(4\).
Common wrong choice: B comes from adding the coefficient and constant in \(4n+1\).
8. Which sequence is an AP for all positive integers \(n\)?
Tested fundamental: Recognising an AP from the general term.
Explanation: A sequence is an AP if \(u_{n+1}-u_n\) is constant. For \(u_n=5-3n\), the common difference is \(-3\).
Common wrong choice: B, because students may see a formula in \(n\) and assume it must be an AP.
Arithmetic Progression: Unknowns and Modelling
9. The 3rd term of an AP is \(22\), and the 7th term is \(46\). Find the first term \(a\) and common difference \(d\).
Answer format: Write your answer as a=...,d=....
Answer: \(a=10,\ d=6\)
Tested fundamental: Forming simultaneous equations using \(u_n=a+(n-1)d\).
Explanation: \(a+2d=22\) and \(a+6d=46\). Subtracting gives \(4d=24\), so \(d=6\). Then \(a=10\).
Common error: Using \(a+3d\) for the 3rd term instead of \(a+2d\).
10. An AP has \(S_5=55\) and \(u_8=31\). Find the first term \(a\) and common difference \(d\).
Answer format: Write your answer as a=...,d=....
Answer: \(a=3,\ d=4\)
Tested fundamental: Forming equations using both a sum and a term.
Explanation: Since \(S_5=55\), \(\frac{5}{2}[2a+4d]=55\), so \(a+2d=11\). Also, \(u_8=31\), so \(a+7d=31\). Hence \(5d=20\), so \(d=4\) and \(a=3\).
Common error: Treating \(S_5\) as the 5th term instead of the sum of the first 5 terms.
11. A savings plan pays a bonus of \(\$20\) in the first month. The bonus increases by \(\$8\) each month. Which expression gives the total bonus over the first \(n\) months?
Tested fundamental: Modelling a real-life AP sum.
Explanation: The bonuses form an AP with \(a=20\) and \(d=8\). Hence \(S_n=\frac{n}{2}[2(20)+(n-1)8]=4n(n+4)\).
Common wrong choice: A gives only the \(n\)th bonus, not the total bonus.
12. A student repays a loan using monthly payments of \[ 100,\ 120,\ 140,\ldots \] Find the least number of payments needed so that the total paid is at least \(780\).
Answer format: Give your answer as an integer.
Answer: \(6\)
Tested fundamental: AP sum with an inequality and integer interpretation.
Explanation: The payments form an AP with \(a=100\) and \(d=20\). Hence \(S_n=\frac{n}{2}[2(100)+(n-1)20]=10n(n+9)\). When \(n=5\), the total is \(700\). When \(n=6\), the total is \(900\). Hence the least number is \(6\).
Common error: Solving the inequality but not checking the least integer value of \(n\).
Geometric Progression: Terms, Sums and Recognition
13. The first term of a GP is \(3\), and the common ratio is \(-2\). Which expression gives the \(n\)th term?
Tested fundamental: General term of a GP.
Explanation: For a GP, \(u_n=ar^{n-1}\). Hence \(u_n=3(-2)^{n-1}\).
Common wrong choice: A comes from using \(r^n\) instead of \(r^{n-1}\).
14. A GP has first term \(2\) and common ratio \(3\). Find the sum of the first \(5\) terms.
Answer format: Give your answer as an integer.
Answer: \(242\)
Tested fundamental: Sum of the first \(n\) terms of a GP.
Explanation: \(S_5=\frac{2(3^5-1)}{3-1}=242\).
Common error: Using the AP sum formula instead of the GP sum formula.
15. True or false: The formula \[ S_n=\frac{a(1-r^n)}{1-r} \] can be used without modification when \(r=1\).
Tested fundamental: Condition for using the finite GP sum formula.
Explanation: When \(r=1\), the denominator \(1-r\) is zero. In this case, all terms are equal to \(a\), so \(S_n=na\).
Common wrong choice: A, from applying the formula without checking \(r\neq1\).
16. Given \[ S_n=7(2^n-1), \] which expression gives the \(n\)th term \(u_n\)?
Tested fundamental: Finding \(u_n\) from \(S_n\) for a GP.
Explanation: \(u_n=S_n-S_{n-1}\). So \(u_n=7(2^n-1)-7(2^{n-1}-1)=7(2^{n-1})\).
Common wrong choice: B, from treating \(S_n\) as though it were already \(u_n\).
17. Which condition is sufficient to show that a sequence is a GP?
Tested fundamental: Showing that a sequence is a GP.
Explanation: A sequence is a GP if the ratio between consecutive terms is constant and non-zero.
Common wrong choice: A is the condition for an AP, not a GP.
18. A GP has \(u_3=20\) and \(u_5=80\). Given that the common ratio is positive, find the common ratio.
Answer format: Give your answer as an integer or improper fraction.
Answer: \(2\)
Tested fundamental: Finding the common ratio from two terms of a GP.
Explanation: \(\frac{u_5}{u_3}=r^2=\frac{80}{20}=4\). Since \(r>0\), \(r=2\).
Common error: Giving \(r=\pm2\) even though the question states that \(r\) is positive.
19. A GP has common ratio \(r\), where \(r\neq0\). What is the common ratio of the sequence formed by its odd-numbered terms \(u_1,u_3,u_5,\ldots\)?
Tested fundamental: Recognising subsequences of a GP.
Explanation: Moving from \(u_1\) to \(u_3\), or from \(u_3\) to \(u_5\), skips one term each time. The multiplier is therefore \(r^2\).
Common wrong choice: A, from assuming the common ratio stays as \(r\) even after skipping terms.
Convergence and Sum to Infinity
20. The geometric series \[ 5+5(k-1)+5(k-1)^2+\cdots \] is convergent when:
Tested fundamental: Finding the range of values for convergence.
Explanation: The common ratio is \(k-1\). For convergence, \(|k-1|<1\). So \(-1<k-1<1\), giving \(0<k<2\).
Common wrong choice: D comes from applying \(|k|<2\) instead of \(|k-1|<1\).
21. A GP has first term \(6\) and common ratio \(\frac{1}{5}\). Find its sum to infinity.
Answer format: Give your answer as an improper fraction.
Answer: \(\frac{15}{2}\)
Tested fundamental: Sum to infinity of a convergent GP.
Explanation: Since \(\left|\frac{1}{5}\right|<1\), \(S_\infty=\frac{a}{1-r}=\frac{6}{1-\frac{1}{5}}=\frac{15}{2}\).
Common error: Using \(1+r\) in the denominator instead of \(1-r\).
22. The infinite geometric series \[ 4+4p+4p^2+\cdots \] has sum to infinity \(10\). Find \(p\).
Answer format: Give your answer as a single fraction in the form p/q.
Answer: \(\frac{3}{5}\)
Tested fundamental: Using \(S_\infty=\frac{a}{1-r}\) to solve for the common ratio.
Explanation: \(\frac{4}{1-p}=10\). So \(1-p=\frac{2}{5}\), and \(p=\frac{3}{5}\).
Common error: Reversing the fraction when solving for \(1-p\).
23. True or false: If a GP has \(|r|<1\), then its terms tend to \(0\), and the infinite geometric series has a finite sum.
Tested fundamental: Link between \(|r|<1\), convergence, and sum to infinity.
Explanation: When \(|r|<1\), \(r^n\to0\) as \(n\to\infty\). This is why the sum to infinity exists.
Common wrong choice: B, from thinking that a negative common ratio must always make the series divergent.
24. Which geometric series is divergent?
Tested fundamental: Identifying when an infinite GP diverges.
Explanation: For \(4-4+4-4+\cdots\), the common ratio is \(-1\). Since \(|r|=1\), the series does not have a sum to infinity.
Common wrong choice: B, because the signs alternate. However, its common ratio is \(-\frac{1}{2}\), so it converges.
Mixed AP/GP Problems and Inequalities
25. Two sequences \((u_n)\) and \((v_n)\) are GPs with positive terms. Their common ratios are \(r\) and \(\frac{1}{R}\) respectively, where \(0<r<R<1\). For the sequence with \(n\)th term \(\ln(u_nv_n)\), what is the common difference?
Tested fundamental: Converting a GP product into an AP using logarithms.
Explanation: \(\ln(u_nv_n)-\ln(u_{n-1}v_{n-1})=\ln\left(\frac{u_n}{u_{n-1}}\cdot\frac{v_n}{v_{n-1}}\right)=\ln\left(r\cdot\frac{1}{R}\right)=\ln\left(\frac{r}{R}\right)\).
Common wrong choice: B reverses the ratio and gives the opposite sign.
26. An account has balance \(B_{n-1}\) at the end of month \(n-1\). At the start of month \(n\), \(\$100\) is withdrawn. At the end of the month, the remaining balance grows by \(2\%\). Which recurrence gives \(B_n\)?
Tested fundamental: Timeline modelling for GP-style financial situations.
Explanation: The withdrawal happens before the interest is applied. So the balance first becomes \(B_{n-1}-100\), then it is multiplied by \(1.02\).
Common wrong choice: A applies interest before subtracting the withdrawal.
27. For a GP with first term \(5\) and common ratio \(\frac{1}{2}\), find the least \(n\) such that \[ S_\infty-S_n<\frac{1}{16}. \]
Answer format: Give your answer as an integer.
Answer: \(8\)
Tested fundamental: Using the remainder after the first \(n\) terms of a convergent GP.
Explanation: \(S_\infty-S_n=\frac{ar^n}{1-r}\). So \(S_\infty-S_n=\frac{5\left(\frac12\right)^n}{1-\frac12}=10\left(\frac12\right)^n\). We need \(10\left(\frac12\right)^n<\frac{1}{16}\), so \(2^n>160\). The least integer is \(n=8\).
Common error: Using \(S_n\) instead of \(S_\infty-S_n\).
28. A drug has a half-life of \(4\) hours. What fraction of a dose remains after \(12\) hours?
Answer format: Give your answer as a single fraction in the form p/q.
Answer: \(\frac{1}{8}\)
Tested fundamental: Modelling repeated percentage decay as a GP.
Explanation: \(12\) hours is \(3\) half-lives. The fraction remaining is \(\left(\frac12\right)^3=\frac18\).
Common error: Applying only one half-life instead of three.
Arithmetic and Geometric Means
29. True or false: The geometric mean between two positive numbers \(a\) and \(b\) is always \(\frac{a+b}{2}\).
Tested fundamental: Difference between arithmetic mean and geometric mean.
Explanation: The arithmetic mean is \(\frac{a+b}{2}\). The geometric mean is \(\sqrt{ab}\) for positive \(a\) and \(b\).
Common wrong choice: A, from confusing AM and GM.
30. Find the positive geometric mean between \(8\) and \(18\).
Answer format: Give your answer as an integer or exact surd.
Answer: \(12\)
Tested fundamental: Finding the geometric mean.
Explanation: The positive geometric mean is \(\sqrt{8\cdot18}=\sqrt{144}=12\).
Common error: Finding the arithmetic mean, \(\frac{8+18}{2}=13\), instead of the geometric mean.
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