Cambridge Additional Mathematics 0606 : 2023 October-November Paper 23 Q10 (Solution)

Question 10(b)

Steps of Solution

A.P.:
$\begin{aligned} a & =3 \\ T_2 & =3+(2-1) d \\ & =3+d \\ T_6 & =3+(6-1) d \\ & =3+5 d \end{aligned}$

G.P.:
$\begin{aligned} a & =3 \\ T_3 & =(3) r^{3-1} \\ & =3 r^2 \\ T_5 & =(3) r^{5-1} \\ & =3 r^4 \\ 3+d & =3 r^2 \\ r^2 & =\frac{3+d}{3} \end{aligned}$

$\begin{aligned} & 3+5 d=3 r^4 =3\left(r^2\right)^2 \\ & 3+5 d=3\left(\frac{3+d}{3}\right)^2 \\ & 3+5 d=\frac{3\left(9+6 d+d^2\right)}{9} \\ & 3(3+5 d)=9+6 d+d^2 \\ & d^2+6 d-15 d+9-9=0 \\ & d^2-9 d=0 \\ & d(d-9)=0 \\ & d=9 \\ & \end{aligned}$

Ignore $d=0$
$\begin{aligned} r^2 & =\frac{3+(9)}{3} \\ r^2 & =4 \\ r & =2 \end{aligned}$

Explanation of the Steps

Arithmetic Progression (A.P.):

1. First term of the A.P.:
$a = 3$

2. Second term of the A.P. ($T_2$):
$T_2 = 3 + (2-1)d$
$T_2 = 3 + d$

3. Sixth term of the A.P. ($T_6$):
$T_6 = 3 + (6-1)d$
$T_6 = 3 + 5d$

Geometric Progression (G.P.):

1. First term of the G.P.:
$a = 3$

2. Third term of the G.P. ($T_3$):
$T_3 = 3 \cdot r^{3-1}$
$T_3 = 3r^2$

3. Fifth term of the G.P. ($T_5$):
$T_5 = 3 \cdot r^{5-1}$
$T_5 = 3r^4$

4. Equating the second term of the A.P. and the third term of the G.P.:
$3 + d = 3r^2$
$r^2 = \frac{3 + d}{3}$

Solving for the Common Difference ($d$) and Common Ratio ($r$):

1. Equating the sixth term of the A.P. and the fifth term of the G.P.:
$3 + 5d = 3r^4$
$3 + 5d = 3(r^2)^2$

2. Substituting $r^2$ from earlier:
$3 + 5d = 3 \left( \frac{3 + d}{3} \right)^2$
$3 + 5d = 3 \cdot \frac{(3 + d)^2}{9}$
$3 + 5d = \frac{3(9 + 6d + d^2)}{9}$

3. Simplifying the equation:
$3(3 + 5d) = 9 + 6d + d^2$
$9 + 15d = 9 + 6d + d^2$
$d^2 + 6d – 15d + 9 – 9 = 0$
$d^2 – 9d = 0$

4. Factoring the quadratic equation:
$d(d – 9) = 0$
$d = 9$

(Ignore $d = 0$ as it is not valid in this context)

5. Substituting $d = 9$ back to find $r$:
$r^2 = \frac{3 + 9}{3}$
$r^2 = \frac{12}{3}$
$r^2 = 4$
$r = 2$

(Ignore $r = -2$ as the common ratio must be greater than 1)

Therefore, the common difference $d$ of the A.P. is 9, and the common ratio $r$ of the G.P. is 2.

ignore $\quad r=-2$