IB DP Maths Topic 1.6 :Simple deductive proof, numerical and algebraic HL Paper 1

Solution:-

(a) attempt to find a difference

$d=p-a, 2d=q-a, d=q-p$ OR $p=a+d, q=a+2d, q=p+d$

correct equation

$p-a=q-p$ OR $q-a=2(p-a)$ OR $p=\frac{a+q}{2}$ (or equivalent)

$2p-q=a$

(b) attempt to find a ratio

$r=\frac{s}{a}, r^{2}=\frac{t}{a}, r=\frac{t}{s}$ OR $s=ar, t=ar^{2}, t=sr$

correct equation

$(\frac{s}{a})^{2}=\frac{t}{a}$ OR $\frac{s}{t}=\frac{t}{s}$ (or equivalent)

$s^{2}=at$

(c) EITHER

$2p – 1 = s^2$ (or equivalent)

$(s^2 > 0) \Rightarrow 2p – 1 > 0$ OR $s = \sqrt{2p-1} \Rightarrow 2p – 1 > 0$ OR $p = \frac{s^2 + 1}{2}$ (and $s^2 > 0$)

OR

$2p – 1 = a$ and $s^2 = a$

$(s^2 > 0, \text{so } a > 0) \Rightarrow 2p – 1 > 0$ OR $p = \frac{a+1}{2}$ and $a > 0$

$\Rightarrow p > \frac{1}{2}$

(d) (i) 9, 5, 1, -3

(ii) 9, 3, 1, $\frac{1}{3}$

(e) (i) attempt to find the difference between two consecutive terms

$d=u_2-u_1=5+\ln3-9-\ln9$ OR $d=u_3-u_2=1+\ln1-5-\ln3$

$\ln9=2\ln3$ OR $\ln1=0$ OR $\ln3-\ln9=\ln\frac{1}{3}(=\ln3^{-1}=-\ln3)$ 

$d=-4-\ln3$

(ii) METHOD 1

attempt to substitute first term and their common difference into $S_{10}$

$\frac{10}{2}(2(9+\ln9)+9(-4-\ln3))$ OR $\frac{10}{2}(2(9+2\ln3)+9(-4-\ln3))$ (or equivalent)

$=5(-18-5\ln3)$ (or equivalent in terms of $\ln3$)

$\sum_{i=1}^{10}u_{i}=-90-25\ln3$

METHOD 2

$u_{10}=9+\ln9+9(-4-\ln3)=(-27+\ln9-9\ln3)$

attempt to substitute first term and their $u_{10}$ into $S_{10}$

$\frac{10}{2}(2(9+\ln9)+9(-4-\ln3))$ OR $\frac{10}{2}(9+\ln9-27+\ln9-9\ln3)$ OR

$\frac{10}{2}(2(9+2\ln3)+9(-4-\ln3))$ OR $\frac{10}{2}(9+\ln9-27-7\ln3)$ (or equivalent)

$=5(-18-5\ln3)$ (or equivalent in terms of $\ln3$)

$\sum_{i=1}^{10}u_{i}=-90-25\ln3$

Solution 

(a) Consider the following, in the arithmetic sequence

Hence proved.

(b) Consider the following in the geometric sequence

Hence proved.

(c) Substitute $q = t = 1$ in

$2p-q=a$   and $s^{2}=at$ , we get

$2p-1=a$  ……….(1)and $s^{2}=a$…………(2), therefore

$2p – 1 = s^2$ …….(3)

Hence proved

(d)(i)  In the arithmetic sequence,

Hence the first four terms of arithmetic sequence are 9, 5, 1, -3

(ii) In the geometric sequence,

First term a = 9,

Hence the first four terms of geometric sequence are 9, 3, 1, $\frac{1}{3}$

(e) (i) finding the common difference,

$d=u_2-u_1=5+\ln3-9-\ln9$ OR $d=u_3-u_2=1+\ln1-5-\ln3$

$\ln9=2\ln3$ OR $\ln1=0$ OR $\ln3-\ln9=\ln\frac{1}{3}(=\ln3^{-1}=-\ln3)$ 

$d=-4-\ln3$

(ii)  The required sum of terms of arithmetic sequence

Hence proved.