CAIE IGCSE Additional Math 12 Series Questions

Covered syllabus points

12.1. Use the binomial theorem 12.2. Use the general binomial term 12.3. Recognise arithmetic and geometric progressions 12.4. Use finite progression formulas 12.5. Use convergence and sum to infinity

[Maximum number: 5]

In this question a, b and n are constants.
When $5(2+a x)^{n}$ is written in ascending powers of x, the first three terms are $640+b^{2} x+30240 x^{2}$ . Find the value of a and the possible values of b.

n=7 soi
$5\times\frac{7\times6}{2}\times2^5\times a^2=30240$ oe
$5\times7\times2^6\times a=b^2$ a=3 only, nfww
$b=\pm\sqrt{6720}$ oe, isw, nfww
If first B1 only scored then SC1 for $a=3\sqrt5$ or $\sqrt{45}$ or 6.71 or 6.708... only, and SC1 for $b=\pm\sqrt{1344\sqrt5}$ or $\pm54.8...$ .

[Maximum number: 6]

Find the exact value of the term independent of x in the expansion of $\left(2+\frac{3}{x^{2}}\right)^{10}\left(1-4 x^{2}\right)^{2}$ .

$2^{10}+10\times2^9\times\frac{3}{x^2}+45\times2^8\times\left(\frac{3}{x^2}\right)^2\ \text{oe, soi}$
and
$1-8x^2+16x^4\ \text{soi}$
B2 for the first expansion;
B1 for two correct terms in the first expansion;
B1 for $1-8x^2+16x^4$ .
$2^{10}+$ their $15360\times$ their (-8)+ their $103680\times$ their 16
M2 FT expansions;
M1 FT for any two correct terms in this sum.
1537024

[Maximum number: 15]

An arithmetic progression, A, has first term a and common difference d. The 2nd, 14th and 17th terms of A form the first three terms of a convergent geometric progression, G, with common ratio r .

(a)

Given that $d\neq0$ , find two expressions for r in terms of a and d and hence show that a=-17d.

[ 6 ]

$a+d,\quad a+13d,\quad a+16d$
soi
$\frac{a+13d}{a+d}=\frac{a+16d}{a+13d}$
oe
$a^2+26ad+169d^2=a^2+17ad+16d^2$ $9ad+153d^2=0$ $9d(a+17d)=0\Rightarrow a=-17d$
oe

(b)

Find the value of r .

[ 2 ]

r=0.25
oe
M1 for $\frac{-17d+13d}{-17d+d}$ or $\frac{-17d+16d}{-17d+13d}$

(c)

The first term of the geometric progression, G , is q and the sum to infinity is $\frac{256}{3}$ . Find the sum of the first 20 terms of the arithmetic progression, A .

[ 7 ]

$\frac{q}{1-0.25}=\frac{256}{3}$
oe
q=64-17d+d=64
oe
$d=-4,\quad a=68$ $S_{20}=600$

[Maximum number: 7]

An arithmetic progression has first term t and common difference 1.5.
The 4th, 8th and 20th terms of this arithmetic progression form the 1st, 2nd and 3rd terms of a geometric progression.

(a)

Find the value of t.

[ 5 ]

At least two correct terms from t+4.5, t+10.5, t+28.5 soi
$r=\frac{t+10.5}{t+4.5}$ , or $r=\frac{t+28.5}{t+10.5}$ , or $r^2=\frac{t+28.5}{t+4.5}$ oe
$\frac{t+10.5}{t+4.5}=\frac{t+28.5}{t+10.5}$ oe
Clears algebraic fractions and expands, e.g. $t^2+21t+110.25=t^2+33t+128.25$ -1.5 oe nfww, and no other values

(b)

Find the common ratio of the geometric progression.

[ 2 ]

3 nfww and no other values
M1 for $\frac{\text{their }(-1.5)+\text{their }10.5}{\text{their }(-1.5)+\text{their }4.5}$ oe, FT their t where valid.
A1

[Maximum number: 5]

A geometric progression, A, has common ratio r, where |r|<1.
The terms of this progression are $a_1, a_2, a_3, \ldots$ .
Another geometric progression, B, has terms $b_1, b_2, b_3, \ldots$ , where
$b_1=a_2,\quad b_2=a_4,\quad b_3=a_6,\ldots$
The sum to infinity of A is $S_A$ and the sum to infinity of B is $S_B$ .
Find $\frac{S_B}{S_A}$ in terms of r.
Give your answer in its simplest form.

$S_A=\frac{a_1}{1-r}$ soi
Allow a for $a_1$ .
$r_b=r^2$ soi
$S_B=\frac{a_2}{1-r^2}=\frac{a_1r}{1-r^2}$
Allow a for $a_1$ with $r^2$ .
$\begin{aligned} k=\frac{S_B}{S_A}&=\frac{a_1r}{1-\text{their }r^2}\times\frac{1-r}{a_1}\\ \frac{S_B}{S_A}&=\frac{\frac{a_1r}{1-\text{their }r^2}}{\frac{a_1}{1-r}} \end{aligned}$
M1 dep on the first and second B1 for their $\frac{S_B}{S_A}$ ;
allow a for $a_1$ .
$\frac{r}{1+r}$ cao

All question bank results loaded

CAIE IGCSE Additional Math 12 Series Topic Practice

Covered syllabus points

Question 1

Question 2

Question 3

Question (a)

Question (b)

Question (c)

Question 4

Question (a)

Question (b)

Question 5