Sequences & series, financial maths, exponents & logs – 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]

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.

4StandardNon-calculator4 marks

Solve $2^{2x}-9\times2^{x}+8=0$.

5StandardNon-calculator5 marks

Solve $\log_2(x+1)+\log_2(x-1)=3$.

Combines logs: $\log_2\big((x+1)(x-1)\big)=3$M1
Converts: $x^2-1=2^3$M1
$x^2=9$ so $x=\pm3$ (or $x=3$ directly, having stated $x>1$)A1
Rejects $x=-3$ since $\log_2(x-1)$ (and $\log_2(x+1)$) is undefined for $x=-3$ (or states the domain $x>1$)R1
$x=3$A1
Answer: $x=3$
Examiner tip: Always check solutions against the domain of the original logarithms.

6HardCalculator7 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 $11.99$.

7HardNon-calculator6 marks

Let $x>0$.

(a) Show that $\log_9 x=\tfrac12\log_3 x$. [2]

(b) Hence, or otherwise, solve $\log_3 x+\log_9 x+\log_{27}x=11$. [4]

Uses change of base: $\log_9x=\dfrac{\log_3x}{\log_39}$M1
$\log_39=2$, so $\log_9x=\tfrac12\log_3x$ AGA1
Writes $\log_{27}x=\tfrac13\log_3x$ (similar method)M1
$\left(1+\tfrac12+\tfrac13\right)\log_3x=\tfrac{11}{6}\log_3x=11$A1
$\log_3x=6$A1
$x=3^6=729$A1
Answer: $x=729$
Examiner tip: In a 'show that' question every step must be visible: state $\log_3 9=2$ explicitly.

Book a free introductory session and we'll find exactly which step is costing you.