Sequences, series and financial applications – IB Maths exam questions with mark scheme

1CoreNon-calculator5 marks

An arithmetic sequence has third term 11 and eighth term 26.

(a) Find the common difference and the first term. [3]

(b) Find the sum of the first 20 terms. [2]

Forms two equations, e.g. $u_1+2d=11$ and $u_1+7d=26$ (or $5d=15$)M1
$d=3$A1
$u_1=5$A1
Uses $S_{20}=\tfrac{20}{2}(2\times5+19\times3)$ or $\tfrac{20}{2}(5+62)$M1
$S_{20}=670$A1
Answer: (a) $d=3$, $u_1=5$ (b) $670$
Examiner tip: Write $u_n=u_1+(n-1)d$ with the right $n$: $u_8$ needs $7d$, not $8d$.

2CoreCalculator4 marks

A geometric sequence begins $6,\ 18,\ 54,\ \dots$

(a) Write down the common ratio and find the 7th term. [2]

(b) Find the sum of the first 8 terms. [2]

$r=3$A1
$u_7=6\times3^6=4374$A1
Uses $S_8=\dfrac{6(3^8-1)}{3-1}$M1
$S_8=19680$A1
Answer: (a) $r=3$, $u_7=4374$ (b) $19680$
Examiner tip: The 7th term uses $r^{6}$, not $r^{7}$.

3StandardCalculator5 marks

Maya invests $\$8000$ in an account that pays a nominal annual interest rate of $3.6\%$, compounded quarterly.

(a) Find the value of the investment after 6 years, giving your answer to the nearest dollar. [2]

(b) Find the least number of complete years after which the value of the investment first exceeds $\$12\,000$. [3]

Substitutes into compound interest formula: $8000\left(1+\dfrac{3.6}{4\times100}\right)^{4\times6}$M1
$\$9919$ (accept $\$9919.23$; accept TVM solver with $N=24$, $I\%=3.6$, $C/Y=4$)A1
Sets up $8000(1.009)^{4n}>12000$ (or $=12000$)M1
$n=11.3$ ($11.31\ldots$) by GDC/TVM or logs (or checks values: after 11 years $\$11\,866$, after 12 years $\$12\,299$)A1
$n=12$ yearsA1
Answer: (a) $\$9919$ (b) $12$ years
Examiner tip: Quarterly compounding means rate $\div4$ and $4n$ periods. Round up in (b) because the question asks for complete years.

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4HardCalculator7 marks

An infinite geometric series has sum to infinity $12$ and second term $3$.

(a) Find the first term and the common ratio. [4]

(b) Find the least value of $n$ for which the sum of the first $n$ terms is greater than $11.99$. [3]

Forms $\dfrac{a}{1-r}=12$ and $ar=3$M1
Eliminates $a$: $12r(1-r)=3$ leading to $4r^2-4r+1=0$M1
$(2r-1)^2=0$, so $r=\tfrac12$A1
$a=6$A1
Uses $S_n=12\left(1-(\tfrac12)^n\right)>11.99$M1
Solves $(\tfrac12)^n<\tfrac{0.01}{12}$ (logs or GDC): $n>10.2$M1
$n=11$A1
Answer: (a) $a=6$, $r=\tfrac12$ (b) $n=11$
Examiner tip: In (b) round up: $n=10$ gives $11.988$, which is not greater than $11.99$.

5CoreNon-calculator5 marks

Let $p = 4.8\times10^{7}$ and $q = 1.6\times10^{-3}$.

(a) Write the number $0.000\,030\,6$ in the form $a\times10^{k}$, where $1\le a<10$ and $k\in\mathbb{Z}$. [1]

(b) Find $pq$, giving your answer in the form $a\times10^{k}$, where $1\le a<10$ and $k\in\mathbb{Z}$. [2]

(a) $3.06\times10^{-5}$A1
(b) Multiplies the numbers and adds the powers: $4.8\times1.6\times10^{7+(-3)}$ (or $7.68\times10^{4}$ seen, or $76\,800$ seen)M1
$7.68\times10^{4}$ (do not accept $76\,800$ or $76.8\times10^{3}$ as the final answer)A1
(c) Divides the numbers and subtracts the powers: $\dfrac{4.8}{1.6}\times10^{7-(-3)}$ (or $3\times10^{10}$ seen)M1
$3\times10^{10}$ (accept $3.0\times10^{10}$)A1
Answer: (a) $3.06\times10^{-5}$ (b) $7.68\times10^{4}$ (c) $3\times10^{10}$
Examiner tip: The A1 marks need the exact form $a\times10^{k}$ with $1\le a<10$; a correct ordinary number such as $76\,800$ loses the final mark, and in (c) dividing by $10^{-3}$ means adding 3 to the power.

6CoreNon-calculator6 marks

Consider the series $\displaystyle\sum_{r=1}^{n}(5r-2)$.

(a) Write down the first three terms of the series. [1]

(b) Show that $\displaystyle\sum_{r=1}^{n}(5r-2)=\frac{n(5n+1)}{2}$. [2]

(a) $3,\ 8,\ 13$A1
(b) Recognises arithmetic with $u_1=3$, $d=5$ and substitutes into $S_n=\frac{n}{2}\left(2u_1+(n-1)d\right)$: $\frac{n}{2}\left(6+5(n-1)\right)$ (or $\frac{n}{2}(u_1+u_n)=\frac{n}{2}\left(3+5n-2\right)$)M1
Simplifies correctly to $\dfrac{n(5n+1)}{2}$ (AG: must see the bracket simplified, e.g. $\frac{n}{2}(5n+1)$, before the given answer)A1
(c) Sets $\dfrac{n(5n+1)}{2}=1010$ and forms the quadratic $5n^2+n-2020=0$M1
Correct attempt to solve: factorises $(n-20)(5n+101)=0$ or uses the quadratic formulaM1
$n=20$ (must reject $n=-\frac{101}{5}$, or give only $n=20$)A1
Answer: (a) $3, 8, 13$ (b) shown (c) $n=20$
Examiner tip: In a 'show that' the final line must follow from working you have written down; and in (c) $n$ counts terms, so the negative root must be rejected.

7StandardCalculator7 marks

A new stand is being built at a sports ground. The diagram shows the first three rows of seats. Row 1, at the front, has 24 seats, and each row has 3 more seats than the row in front of it.

(a) Find the number of seats in row 15. [2]

(b) Find the total number of seats in the first 15 rows. [2]

The stand must hold at least 2000 seats.

(a) Uses $u_n=u_1+(n-1)d$: $24+14\times3$M1
$66$A1
(b) Uses an arithmetic series formula: $\frac{15}{2}(24+66)$ or $\frac{15}{2}(2\times24+14\times3)$M1
$675$A1
(c) Sets up $\frac{n}{2}\left(48+3(n-1)\right)\ge2000$ (accept $=2000$ or $>2000$)M1
$n=29.78\ldots$ from GDC (or checks $S_{29}=1914$ and $S_{30}=2025$)A1
$30$ rowsA1
Answer: (a) $66$ (b) $675$ (c) $30$ rows
Examiner tip: Use $n-1$ (not $n$) lots of $d$ in $u_n$, and in (c) round the non-integer solution up, because 29 rows only hold 1914 seats.

8StandardCalculator8 marks

A company buys a machine, machine A, for €45 000. The value of machine A depreciates by $12\%$ per year.

(a) Find the value of machine A after 5 years. [2]

(b) Find the number of complete years after which the value of machine A first falls below €15 000. [3]

The company also buys machine B for €45 000. The value of machine B decreases by €3600 each year.

(a) $45000\times0.88^{5}$ (accept $45000(1-0.12)^5$)M1
€$23\,700$ (€$23\,747.94$; accept €$23\,748$)A1
(b) Sets up $45000\times0.88^{n}<15000$ (accept $=$)M1
$n=8.59\ldots$ by GDC or logs (or values seen: after 8 years €$16\,184$, after 9 years €$14\,242$)A1
$9$ yearsA1
(c) Value of A after 8 years: $45000\times0.88^{8}=$ €$16\,183.55$ (accept €$16\,184$)A1
Value of B after 8 years: $45000-8\times3600=$ €$16\,200$A1
Machine B has the greater value, with both values seen and compared (R1 depends on both values correct; do not accept 3 s.f. values for A, since $16\,200$ vs $16\,200$ cannot decide)R1
Answer: (a) €$23\,700$ (3 s.f.) (b) $9$ years (c) Machine B (€$16\,200$ against €$16\,183.55$)
Examiner tip: Depreciation of $12\%$ means multiplying by $0.88$ each year, not $0.12$. In (c), A to 3 s.f. ties with B, so keep more figures before deciding.

9HardCalculator8 marks

At the start of each year, Kofi pays $\$500$ into a savings account. The account pays compound interest at $4\%$ per year, added at the end of each year. He makes no withdrawals.

(a) Show that the amount in the account at the end of year 3 is $500\left(1.04+1.04^{2}+1.04^{3}\right)$. [2]

(b) Find the amount in the account at the end of year 10. Give your answer to the nearest dollar. [3]

(a) States that the first payment earns interest for 3 years, the second for 2 years and the third for 1 year, e.g. $500\times1.04^{3}$, $500\times1.04^{2}$, $500\times1.04$M1
Adds the three amounts to give $500\left(1.04+1.04^{2}+1.04^{3}\right)$ (AG; accept a year-by-year build-up $((500\times1.04+500)\times1.04+500)\times1.04$ expanded to the given form)A1
(b) Recognises a geometric series with first term $520$ (or $500\times1.04$) and common ratio $1.04$M1
Substitutes into $S_n$ with $n=10$: $\dfrac{520\left(1.04^{10}-1\right)}{1.04-1}$M1
$\$6243$ (A0 if the final answer is left as $\$6243.18$, since the nearest dollar is required; award A1 if $6243.18$ is seen and then correctly rounded)A1
(c) Sets up $\dfrac{520\left(1.04^{n}-1\right)}{0.04}>10000$ (accept $=$)M1
$n=14.5\ldots$ by GDC or logs (or values seen: $S_{14}=\$9512$, $S_{15}=\$10\,412$)A1
$15$ yearsA1
Answer: (a) shown (b) $\$6243$ (c) $15$ years
Examiner tip: Because the money goes in at the start of each year, every payment earns at least one year of interest, so the first term is $520$, not $500$.

10StretchNon-calculator8 marks

The diagram shows a sequence of squares $S_1, S_2, S_3, \dots$. Square $S_1$ has side length $8\text{ cm}$. Each square after $S_1$ is formed by joining the midpoints of the sides of the previous square. The sequence continues indefinitely.

(a) Show that the side lengths of the squares form a geometric sequence with common ratio $\dfrac{1}{\sqrt{2}}$. [2]

(b) Show that the sum of the perimeters of all the squares is $(64+32\sqrt{2})\text{ cm}$. [4]

(a) Uses Pythagoras on a corner triangle with legs $\frac{x}{2}$, $\frac{x}{2}$: next side $=\sqrt{\left(\frac{x}{2}\right)^2+\left(\frac{x}{2}\right)^2}$ (or $\sqrt{4^2+4^2}=4\sqrt{2}$ for $S_2$)M1
$=\dfrac{x}{\sqrt{2}}$, so each side is $\frac{1}{\sqrt{2}}$ times the previous one, a constant ratio (AG; accept showing $\frac{4\sqrt{2}}{8}=\frac{4}{4\sqrt{2}}=\frac{1}{\sqrt{2}}$ with the general argument stated)A1
(b) Perimeters form a geometric sequence with first term $32$ and ratio $\frac{1}{\sqrt{2}}$; notes $\left|\frac{1}{\sqrt{2}}\right|<1$ so the sum to infinity existsR1
Substitutes into $S_\infty=\dfrac{u_1}{1-r}$: $\dfrac{32}{1-\frac{1}{\sqrt{2}}}$ (or equivalent, e.g. $\dfrac{32\sqrt{2}}{\sqrt{2}-1}$)M1
Rationalises the denominator: $\dfrac{32\sqrt{2}(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}$M1
$=\dfrac{64+32\sqrt{2}}{1}=64+32\sqrt{2}$ (AG; the denominator $2-1=1$ must be seen or implied)A1
(c) Areas form a geometric sequence with ratio $\left(\frac{1}{\sqrt{2}}\right)^2=\frac{1}{2}$ (accept $64, 32, 16,\dots$)M1
$S_\infty=\dfrac{64}{1-\frac{1}{2}}=128=2\times64$, i.e. twice the area of $S_1$ (accept equivalent general argument $\dfrac{A}{1-\frac12}=2A$)A1
Answer: (a) shown (b) $(64+32\sqrt{2})\text{ cm}$ (c) area ratio $\frac12$, so $S_\infty=\frac{64}{1/2}=128=2\times64\text{ cm}^2$
Examiner tip: The ratio for areas is the square of the ratio for lengths, so it is $\frac12$, not $\frac{1}{\sqrt2}$; and state $|r|<1$ before you use the sum to infinity.

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