Comprehensive assessment for the Sequences and Series focus area: arithmetic sequences and series, geometric sequences and series, sigma notation, the limiting sum of a geometric progression, recurrence relations, and financial applications such as compound interest and depreciation.
Assessment
Select the best answer for each question. 1 mark each.
An arithmetic sequence has first term $3$ and common difference $4$. The $10$th term $T_{10}$ is:
An arithmetic sequence has first term $5$ and common difference $-3$. The $8$th term $T_8$ is:
For the arithmetic series with first term $2$ and common difference $3$, the sum of the first $20$ terms $S_{20}$ is:
Which of the following is a geometric sequence?
A geometric sequence has first term $3$ and common ratio $2$. The $6$th term $T_6$ is:
For the geometric series with first term $1$ and common ratio $2$, the sum of the first $6$ terms $S_6$ is:
The limiting sum of the geometric series $8 + 4 + 2 + 1 + \ldots$ is:
A geometric series has a limiting sum if and only if the common ratio $r$ satisfies:
Evaluate $\displaystyle\sum_{n=1}^{4} (2n + 1)$.
An amount is invested at $6\%$ per annum compound interest. The year-end balances form a geometric progression whose common ratio is:
A geometric sequence has first term $5$ and fourth term $40$. The common ratio $r$ is:
For the arithmetic sequence $7,\ 10,\ 13,\ \ldots$, which term is equal to $100$?
A sequence is defined by the recurrence $T_{n+1} = T_n + 5$ with $T_1 = 2$. The value of $T_3$ is:
The limiting sum of the geometric series $0.4 + 0.04 + 0.004 + \ldots$ is:
A machine bought for $\$20{,}000$ depreciates by $10\%$ of its value each year (reducing balance). Its value at the end of $3$ years is:
Short Answer
An arithmetic sequence has first term $a = 4$ and common difference $d = 6$. Find (a) the $12$th term, and (b) the sum of the first $12$ terms. Show all working.
A geometric sequence has first term $a = 5$ and common ratio $r = 3$. Find (a) the $5$th term, and (b) the sum of the first $5$ terms. Show all working.
Find the limiting sum of the geometric series $18 + 12 + 8 + \ldots$, first stating why the limiting sum exists. Show all working.
The $3$rd term of an arithmetic sequence is $11$ and the $7$th term is $27$. Find the first term $a$ and the common difference $d$. Show all working.
Evaluate $\displaystyle\sum_{n=1}^{10} (3n - 1)$ by recognising it as an arithmetic series and using the sum formula. Show all working.
How many terms of the geometric series $2 + 6 + 18 + \ldots$ are needed for the sum to first exceed $2000$? Show all working.
$\$8{,}000$ is invested at $5\%$ per annum compound interest, compounded annually. The year-end balances form a geometric progression. Find (a) the first term $a$ and common ratio $r$ of this GP, and (b) the balance at the end of $4$ years, correct to the nearest dollar. Show all working.
At the start of each year for $10$ years, $\$1{,}000$ is deposited into an account earning $6\%$ per annum compound interest, compounded annually. Find the total value of the annuity at the end of the $10$th year, correct to the nearest dollar. Show all working.
Q1: D, $T_{10} = a + (n-1)d = 3 + 9 \times 4 = 39$.
Q2: B, $T_8 = a + (n-1)d = 5 + 7 \times (-3) = 5 - 21 = -16$.
Q3: C, $S_{20} = \frac{n}{2}\left(2a + (n-1)d\right) = 10\left(2 \times 2 + 19 \times 3\right) = 10(4 + 57) = 610$.
Q4: A, $2, 6, 18, 54, \ldots$ has a constant ratio $\frac{6}{2} = \frac{18}{6} = 3$. Option B is arithmetic ($d = 3$), D is arithmetic ($d = 3$), and C has neither a constant ratio nor a constant difference.
Q5: D, $T_6 = ar^{n-1} = 3 \times 2^5 = 3 \times 32 = 96$.
Q6: C, $S_6 = \frac{a(r^n - 1)}{r - 1} = \frac{1(2^6 - 1)}{2 - 1} = 64 - 1 = 63$.
Q7: B, $r = \frac{4}{8} = \frac{1}{2}$ and $|r| < 1$, so $S_\infty = \frac{a}{1 - r} = \frac{8}{1 - \frac{1}{2}} = \frac{8}{\frac{1}{2}} = 16$.
Q8: A, a geometric series converges to a limiting sum $S_\infty = \frac{a}{1 - r}$ exactly when $|r| < 1$.
Q9: D, $\sum_{n=1}^{4}(2n + 1) = 3 + 5 + 7 + 9 = 24$.
Q10: B, each balance is the previous balance multiplied by $(1 + i) = 1 + 0.06 = 1.06$. The common ratio is $(1 + i)$, never just $i$.
Q11: C, $T_4 = ar^3 = 5r^3 = 40$, so $r^3 = 8$ and $r = 2$.
Q12: D, $a = 7$, $d = 3$, and $T_n = 7 + (n-1)3 = 100$ gives $3(n-1) = 93$, so $n - 1 = 31$ and $n = 32$.
Q13: A, this is an arithmetic sequence with $d = 5$: $T_1 = 2$, $T_2 = 7$, $T_3 = 12$.
Q14: B, $a = 0.4$ and $r = 0.1$, so $S_\infty = \frac{a}{1 - r} = \frac{0.4}{0.9} = \frac{4}{9}$.
Q15: C, reducing balance depreciation gives $V = 20{,}000(1 - 0.10)^3 = 20{,}000(0.9)^3 = 20{,}000 \times 0.729 = 14{,}580$.
Q16 (3 marks): (a) $T_{12} = a + (n-1)d = 4 + 11 \times 6 = 4 + 66 = 70$ [1]. (b) $S_{12} = \frac{n}{2}\left(2a + (n-1)d\right) = \frac{12}{2}\left(2 \times 4 + 11 \times 6\right) = 6(8 + 66) = 6 \times 74 = 444$ [2].
Q17 (3 marks): (a) $T_5 = ar^{n-1} = 5 \times 3^4 = 5 \times 81 = 405$ [1]. (b) $S_5 = \frac{a(r^n - 1)}{r - 1} = \frac{5(3^5 - 1)}{3 - 1} = \frac{5(243 - 1)}{2} = \frac{5 \times 242}{2} = 605$ [2].
Q18 (3 marks): $r = \frac{12}{18} = \frac{2}{3}$ [1]. Since $|r| = \frac{2}{3} < 1$, the limiting sum exists [1]. $S_\infty = \frac{a}{1 - r} = \frac{18}{1 - \frac{2}{3}} = \frac{18}{\frac{1}{3}} = 54$ [1].
Q19 (3 marks): $T_3 = a + 2d = 11$ and $T_7 = a + 6d = 27$ [1]. Subtracting gives $4d = 16$, so $d = 4$ [1]. Then $a + 2(4) = 11$, so $a = 3$ [1].
Q20 (4 marks): The terms are $3n - 1$: at $n = 1$, $a = 2$; the common difference is $d = 3$ [1]. There are $n = 10$ terms, and the last term is $T_{10} = 3(10) - 1 = 29$ [1]. $S_{10} = \frac{n}{2}\left(2a + (n-1)d\right) = \frac{10}{2}\left(2 \times 2 + 9 \times 3\right) = 5(4 + 27) = 5 \times 31 = 155$ [2]. (Check: $\frac{10}{2}(2 + 29) = 5 \times 31 = 155$.)
Q21 (4 marks): $a = 2$, $r = 3$, so $S_n = \frac{2(3^n - 1)}{3 - 1} = 3^n - 1$ [1]. Require $3^n - 1 > 2000$, that is $3^n > 2001$ [1]. Testing powers: $3^6 = 729$ and $3^7 = 2187$ [1]. Since $3^7 = 2187 > 2001$ but $3^6 = 729 < 2001$, the smallest $n$ is $7$. So $7$ terms are needed [1].
Q22 (4 marks): (a) The first year-end balance is $a = P(1 + i) = 8{,}000 \times 1.05 = 8{,}400$, and the common ratio is $r = 1 + i = 1.05$ [2]. (b) The balance after $4$ years is $P(1 + i)^4 = 8{,}000 \times 1.05^4 = 8{,}000 \times 1.21550625 = 9{,}724.05$, which is $\$9{,}724$ to the nearest dollar [2].
Q23 (4 marks): Each $\$1{,}000$ deposit is made at the start of a year, so it earns interest for a whole number of years by the end of year $10$. The first deposit grows for $10$ years to $1{,}000(1.06)^{10}$, and the last deposit grows for $1$ year to $1{,}000(1.06)^{1}$ [1]. These form a GP with first term $a = 1{,}000(1.06) = 1{,}060$ and common ratio $r = 1.06$, with $n = 10$ terms [1]. $A = \frac{a(r^n - 1)}{r - 1} = \frac{1{,}060\left(1.06^{10} - 1\right)}{0.06}$ [1]. Since $1.06^{10} = 1.790847$, $A = \frac{1{,}060 \times 0.790847}{0.06} = \frac{838.30}{0.06} = 13{,}971.64$, which is $\$13{,}972$ to the nearest dollar [1].