Quadratic Functions (IGCSE Old Questions)

1. $f(x)=x^{2}+6 x+8$
   
   Given that $f(x)$ can be expressed in the form $(x+A)^{2}+B$ where $A$ and $B$ are constants,
   1. find the value of $A$ and the value of $B$.
   2. Hence, or otherwise, find
      1. the value of $x$ for which $f(x)$ has its least value
      2. the least value of $f(x)$
   The curve $C$ has equation $y=x^{2}+6 x+8$ The line $l$, with equation $y=2-x$, intersects $C$ at two points.
   3. Find the $x$-coordinate of each of these two points.
   4. Find the $x$-coordinate of the points where $C$ crosses the $x$-axis.



$\begin{aligned} f(x) &=x^{2}+6 x+8 \\\\ &=x^{2}+2(3 x)+9-1 \\\\ &=\left(x^{2}+3\right)^{2}-1\\\\ f(x) &=x^{2}+6 x+8 \\\\ &=x^{2}+2(3 x)+9-1 \\\\ &=\left(x^{2}+3\right)^{2}-1 \end{aligned}$ (a) $A=3$ and $B=-1$ (b) (i) $f(x)$ has the least value when $x=-3$. $\quad$(ii) The least value of $f(x)=-1$. $\begin{aligned} & C: \quad y=x^{2}+6 x+8 \\\\ & l: \quad y=2-x \end{aligned}$ (c) At the points of intersection of $C$ and $l$, $\begin{aligned} & x^{2}+6 x+8=2-x \\\\ &\therefore x^{2}+7 x+6=0 \\\\ &(x+6)(x+1)=0 \\\\ &\therefore x=-6 \text { or } x=-1 \\\\ &\text { At } x=-6, y=2-(-6)=8 \\\\ &\text { At } x=-1, y=2-(-1)=3 \end{aligned}$ $\therefore\quad$ The point of intersection of $C$ and $l$ are $(-6,8)$ and $(-1,3)$. (d) When $C$ cross $x$-axis, $\begin{aligned} & x^{2}+6 x+8=0 \\\\ & (x+2)(x+4)=0 \end{aligned}$ $\therefore\quad$ The curve $C$ cross $x$-axis at $(-2,0)$ and $(-4,0)$.




2. $f(x)=3 x^{2}+6 x+7$
   
   Given that $\mathrm{f}(x)$ can be written in the form $A(x+B)^{2}+C$, where $A, B$ and $C$ are rational numbers,
   1. find the value of $A$, the value of $B$ and the value of $C$.
   2. Hence, or otherwise, find
      1. the value of $x$ for which $\dfrac{1}{f(x)}$ is a maximum,
      2. the maximum value of $\dfrac{1}{f(x)}$.



$\begin{aligned} f(x) &=3 x^{2}+6 x+7 \\\\ &=3 x^{2}+6 x+3+4 \\\\ &=3\left(x^{2}+2 x+1\right)+4 \\\\ &=3(x+1)^{2}+4 \end{aligned}$ (a) $A=3, B=1, C=4$ (b) (i) $\dfrac{1}{f(x)}$ is maximum when $f(x)$ is minimun. $\quad \quad f(x)$ is minimun when $x=-1.$ $\therefore \quad\dfrac{1}{f(x)}$ is maximum when $x=-1.$ $\quad$ (ii) The minimum value of $f(x)$ is 4. $\therefore \quad$ The maximum value of $f(x)$ is $\dfrac{1}{4}$.




3. $f(x)=2 x^{2}-8 x+5$
   
   Given that $f(x)$ can be written in the form $a(x-b)^{2}+c$
   1. find the value of $a$, the value of $b$ and the value of $c$.
   2. Write down
      1. the minimum value of $f(x)$
      2. the value of $x$ at which this minimum occurs.



$\begin{aligned} f(x) &=2 x^{2}-8 x+5 \\\\ &=2\left(x^{2}-4 x\right)+5 \\\\ &=2\left(x^{2}-4 x+4\right)+5-8 \\\\ &=2(x-2)^{2}-3 \end{aligned}$ (a) $a=2, b=2$ and $c=-3$ (b) (i) the minimun value of $f(x)=-3$ $\quad$ (ii) $f(x)$ is minimum when $x=2$.




4. $f(x)=6+5 x-2 x^{2}$
   
   Given that $f(x)$ can be written in the form $p(x+q)^{2}+r$, where $p, q$ and $r$ are rational numbers,
   1. find the value of $p$, the value of $q$ and the value of $r$.
   2. Hence, or otherwise, find
      1. the maximum value of $f(x)$
      2. the value of $x$ for which this maximum occurs.
   $g(x)=6+5 x^{3}-2 x^{6}$
   
   
   3. Write down
      1. the maximum value of $g(x)$,
      2. the exact value of $x$ for which this maximum occurs.



$\begin{aligned} f(x) &=6+5 x-2 x^{2} \\\\ &=-2\left(x^{2}-\dfrac{5}{2} x\right)+6 \\\\ &=-2\left(x^{2}-2\left(\dfrac{5}{4}\right) x+\left(\dfrac{5}{4}\right)^{2}\right)+6+2\left(\dfrac{5}{4}\right)^{2} \\\\ &=-2\left(x-\dfrac{5}{4}\right)^{2}+6+\dfrac{25}{8} \\\\ &=-2\left(x-\dfrac{5}{4}\right)^{2}+\dfrac{73}{8} \end{aligned}$ (a) $P=-2, \quad q=-\dfrac{5}{4}, r=\dfrac{73}{8}$ (b) (i) the maximum value of $f(x)=\dfrac{73}{8}$ $\quad$ (ii) $f(x)$ is maximun when $x=\dfrac{5}{4}$ $\quad\quad g(x)=6+5 x^{3}-2 x^{6}$ $\quad\quad\quad\quad=-2\left(x^{3}-\dfrac{5}{4}\right)^{2}+\dfrac{73}{8}$ (c) the maximum value of $g(x)=\dfrac{73}{8}$ $\quad g(x)$ has maximum value when $\begin{aligned} \quad &x^{3}-\dfrac{5}{4}=0 \\ \quad &x=\sqrt[3]{\dfrac{5}{4}} \end{aligned}$




5. The roots of the equation $x^{2}+6 x+2=0$ are $\alpha$ and $\beta$, where $\alpha>\beta$. Without solving the equation,
   1. find
      1. the value of $\alpha^{2}+\beta^{2}$,
      2. the value of $\alpha^{4}+\beta^{4}$.
   2. Show that $\alpha-\beta=2 \sqrt{7}$.
   3. Factorise completely $\alpha^{4}-\beta^{4}$.
   4. Hence find the exact value of $\alpha^{4}-\beta^{4}$.
   5. Given that $\beta^{4}=A+B \sqrt{7}$ where $A$ and $B$ are positive constants find the value of $A$ and the value of $B$.



$x^{2}+6 x+2=0$ $\alpha$ and $\beta$ are the roots of the equation. $\begin{aligned} \therefore \quad x^{2}+6 x+2 &=(x-\alpha)(x-\beta) \\\\ &=x^{2}-(\alpha+\beta) x+\alpha \beta \\\\ \therefore \quad \alpha+\beta=-6 & \text { and } \alpha \beta=2 \end{aligned}$ $\text{(a) (i) Since we have } (\alpha+\beta)^{2}=\alpha^{2}+2 \alpha \beta+\beta^{2}$, $\begin{aligned} \alpha^{2}+\beta^{2} &=(\alpha+\beta)^{2}-2 \alpha \beta \\\\ &=36-4 \\\\ &=32 \end{aligned}$ $\begin{aligned} \text{(ii)}\qquad \alpha^{4}+\beta^{4} & \\\\ \left(\alpha^{2}\right)^{2}+\left(\beta^{2}\right)^{2}&=\left(\alpha^{2}+\beta^{2}\right)^{2}-2\left(\alpha^{2} \beta^{2}\right) \\\\ &=\left(\alpha^{2}+\beta^{2}\right)^{2}-2(\alpha \beta)^{2} \\\\ &=(32)^{2}-2(2)^{2} \\\\ &=1016 \end{aligned}$ $ \begin{aligned} \text{(b) Since }(\alpha-\beta)^{2} &=\alpha^{2}-2 \alpha \beta+\beta^{2} \\\\ &=\alpha^{2}+\beta^{2}-2(\alpha \beta) \\\\ &=32-2(2) \\\\ &=28 \\\\ \alpha-\beta &=\sqrt{28} \\\\ &=2 \sqrt{7} \end{aligned}$ $\begin{aligned} \text{(c) } \alpha^{4}-\beta^{4} &=\left(\alpha^{2}-\beta^{2}\right)\left(\alpha^{2}+\beta^{2}\right) \\\\ &=(\alpha-\beta)(\alpha+\beta)\left(\alpha^{2}+\beta^{2}\right) \end{aligned}$




6. The equation $x^{2}+m x+15=0$ has roots $\alpha$ and $\beta$ and the equation $x^{2}+h x+k=0$ has roots $\dfrac{\alpha}{\beta}$ and $\dfrac{\beta}{\alpha}$
   1. Write down the value of $k$.
   2. Find an expression for $h$ in terms of $m$.
   
   Given that $\beta=2 \alpha+1$,
   
   
   3. find the two possible values of $\alpha$.
   4. Hence find the two possible values of $m$.



$\alpha$ and $\beta$ are the roots of the equation $x^{2}+m x+15=0$ $\begin{aligned} \therefore \quad & x^{2}+m x+15=(x-\alpha)(x-\beta) \\\\ & x^{2}+m x+15=x^{2}-(\alpha+\beta) x+\alpha \beta \end{aligned}$ $\therefore \alpha+\beta=-m$ and $\alpha \beta=15$ $\dfrac{\alpha}{\beta}$ and $\dfrac{\beta}{\alpha}$ are the rodts of the equation $x^{2}+h x+k=0$ $\therefore x^{2}+h x+k=\left(x-\dfrac{\alpha}{\beta}\right)\left(x-\dfrac{\beta}{\alpha}\right)$ $\therefore x^{2}+h x+k=x^{2}-\left(\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}\right) x+1$ $\begin{aligned} \text {(a) } k&=1\\\\ \text { (b) }h &=-\left(\dfrac{\alpha}{\beta}+\dfrac{B}{\alpha}\right) \\\\ &=-\dfrac{\alpha^{2}+\beta^{2}}{\alpha \beta} \\\\ &=-\dfrac{(\alpha+\beta)^{2}-2 \alpha \beta}{d \beta} \\\\ &=-\dfrac{m^{2}-30}{15} \\\\ &=\dfrac{30-m^{2}}{15} \end{aligned}$ $\begin{aligned} \text { (c) }&\beta=2 \alpha+1 \\\\ &\alpha \beta=15 \\\\ & \alpha(2 d+1)=15 \\\\ &2 \alpha^{2}+\alpha-15=0 \\\\ &(\alpha+3)(2 \alpha-5)=0 \\\\ &\alpha=-3 \text { or } \alpha=\dfrac{5}{2} \end{aligned}$ $\begin{aligned} \text { (d) } \alpha+\beta &=-m \\\\ m &=-\alpha-\beta \\\\ &=-\alpha-2 \alpha-1 \\\\ &=-(3 \alpha+1) \\\\ \text { when } \alpha &=-3, m=8 \\\\ \text { when } \alpha &=\dfrac{5}{2}, m=-\dfrac{17}{2} \end{aligned}$




7. The equation $2 x^{2}-7 x+4=0$ has roots $\alpha$ and $\beta$ Without solving this equation, form a quadratic equation with integer coefficients which has roots $\alpha+\dfrac{1}{\beta}$ and $\beta+\dfrac{1}{\alpha}$.



The roots of the equation $2 x^{2}-7 x+4=0$ are $\alpha$ and $\beta$. $\begin{aligned} \therefore 2 x^{2}-7 x+4 &=2(x-\alpha)(x-\beta) \\\\ &=2 x^{2}-2(\alpha+\beta) x+2 \alpha \beta \\\\ \therefore \alpha+\beta=\dfrac{7}{2},\ & \alpha \beta=2 \end{aligned}$ The quadratic equation which has roots $\alpha+\dfrac{1}{\beta}$ and $\beta+\dfrac{1}{\alpha}$ is $\begin{aligned} &\left[x-\left(\alpha+\dfrac{1}{\beta}\right)\right]\left[x-\left(\beta+\dfrac{1}{\alpha}\right)\right]=0 \\\\ \therefore\quad & x^{2}-\left(\alpha+\beta+\dfrac{1}{\alpha}+\dfrac{1}{\beta}\right) x+\left(\alpha \beta+\dfrac{1}{\alpha \beta}+2\right)=0 \end{aligned}$ $\begin{aligned} \quad &x^{2}-\left(\alpha+\beta+\dfrac{\alpha+\beta}{\alpha \beta}\right) x+\left(\alpha \beta+\dfrac{1}{\alpha \beta}+2\right)=0 \\\\ \quad &x^{2}-\left(\dfrac{7}{2}+\dfrac{7}{4}\right) x+\left(2+\dfrac{1}{2}+2\right)=0 \\\\ \quad &4 x^{2}-(14+7) x+(8+2+8)=0 \\\\ \quad &4 x^{2}-21 x+18=0 \end{aligned}$




8. $f(x)=2 x^{2}-5 x+1$
   
   The equation $f(x)=0$ has roots $\alpha$ and $\beta$. Without solving the equation
   1. find the value of $\alpha^{2}+\beta^{2}$
   2. show that $\alpha^{4}+\beta^{4}=\dfrac{433}{16}$
   3. form a quadratic equation with integer coefficients which has roots $\left(\alpha^{2}+\dfrac{1}{\alpha^{2}}\right)$ and $\left(\beta^{2}+\dfrac{1}{\beta^{2}}\right)$



$f(x)=2 x^{2}-5 x+1$ The roots of the equation $f(x)=0$ are $\alpha$ and $\beta$ $\begin{aligned} &2 x^{2}-5 x+1=2(x-\alpha)(x-\beta) \\\\ &2 x^{2}-5 x+1=2 x^{2}-2(\alpha+\beta) x+22 \beta \end{aligned}$ $\therefore \alpha+\beta=\dfrac{5}{2}$ and $\alpha \beta=\dfrac{1}{2}$ $ \begin{aligned} \text { (a) } \alpha^{2}+\beta^{2} &=\left(\alpha+\beta^{2}\right)-2 \alpha \beta \\\\ &=\dfrac{25}{4}-1 \\\\ &=\dfrac{21}{4}\\\\ \text { (b) } \alpha^{4}+\beta^{4} &=\left(\alpha^{2}+\beta^{2}\right)^{2}-2(\alpha \beta)^{2} \\\\ &=\left(\dfrac{21}{4}\right)^{2}-2\left(\dfrac{1}{2}\right)^{2} \\\\ &=\dfrac{433}{16} \end{aligned}$ $\text { (c) }$ The quadratic equation which has roots $\left(\alpha^{2}+\dfrac{1}{\alpha^{2}}\right)$ and $\left(\beta^{2}+\dfrac{1}{\beta^{2}}\right)$ is $\begin{aligned} &{\left[x-\left(\alpha^{2}+\dfrac{1}{\alpha^{2}}\right)\right]\left[x-\left(\beta^{2}+\dfrac{1}{\beta^{2}}\right)\right]=0} \\\\ &x^{2}-\left(a^{2}+\beta^{2}+\dfrac{1}{\alpha^{2}}+\dfrac{1}{\beta^{2}}\right) x+\left(\alpha^{2} \beta^{2}+\dfrac{\alpha^{2}}{\beta^{2}}+\dfrac{\beta^{2}}{\alpha^{2}}+\dfrac{1}{\alpha^{2} \beta^{2}}\right)=0 \\\\ &x^{2}-\left[\alpha^{2}+\beta^{2}+\dfrac{\alpha^{2}+\beta^{2}}{(\alpha \beta)^{2}}\right] x+\left((a \beta)^{2}+\dfrac{\alpha^{4}+\beta^{4}}{(\alpha \beta)^{2}}+\dfrac{1}{(\alpha \beta)^{2}}\right)=0 \\\\ &x^{2}-\left(\dfrac{21}{4}+\dfrac{21}{4} \times 4\right) x+\left(\dfrac{1}{4}+\dfrac{433}{16}(4)+4\right)=0 \\\\ &4 x^{2}-105 x+450=0 \end{aligned}$




9. The equation $x^{2}+p x+1=0$ has roots $\alpha$ and $\beta$
   1. Find, in terms of $p$, an expression for
      1. $\alpha+\beta$.
      2. $\alpha^{2}+\beta^{2}$.
      3. $\alpha^{3}+\beta^{3}$.
   2. Find a quadratic equation, with coefficients expressed in terms of $p$, which has roots $a^{3}$ and $\beta^{3}$.



$\therefore \quad x^{2}+p x+1=(x-\alpha)(x-\beta)$ $x^{2}+p x+1=x^{2}-(\alpha+\beta) x+\alpha \beta$ $\begin{aligned} \text{(a) } &\alpha+\beta=-p, \alpha \beta=1\\\\ \text{(i) } & \therefore\ \alpha+\beta=-p\\\\ \text{(ii) }& \alpha^{2}+\beta^{2}\\\\ & =(\alpha+\beta)^{2}-2 \alpha \beta \\\\ &=p^{2}-1\\\\ \text{(iii) } &\text { Since }(\alpha+\beta)^{3}=\alpha^{3}+3 \alpha^{2} \beta+3 \alpha \beta^{2}+\beta^{3} \\\\ &=\alpha^{3}+\beta^{3}+3 \alpha \beta(\alpha+\beta) \\\\ &\alpha^{3}+\beta^{3}=\left(\alpha+\beta^{3}\right)-3 \alpha \beta(\alpha+\beta)\\\\ &\alpha^{3}+\beta^{3}=-p^{3}-3(-p)=3 p-p^{3} \end{aligned}$ $\text{(b)}$ The guadratic equation Which has roots $\alpha^{3}$ and $\beta^{3}$ is $\begin{aligned} &\left(x-\alpha^{3}\right)\left(x-\beta^{3}\right)=0 \\\\ &x^{2}-\left(\alpha^{3}+\beta^{3}\right) x+(\alpha \beta)^{3}=0 \\\\ &x^{2}-\left(3 p-p^{3}\right) x+1=0 \\\\ &x^{2}+\left(p^{3}-3 p\right) x+1=0 \end{aligned}$




10. 1. Show that $(\alpha+\beta)\left(\alpha^{2}-\alpha \beta+\beta^{2}\right)=\alpha^{3}+\beta^{3}$.
       The roots of the equation $2 x^{2}+6 x-7=0$ are $\alpha$ and $\beta$ where $\alpha>\beta$.
       Without solving the equation,
    2. find the value of $\alpha^{3}+\beta^{3}$.
    3. show that $\alpha-\beta=\sqrt{23}$.
    4. Hence find the exact value of $\alpha^{3}-\beta^{3}$.



$\begin{aligned} \text{(a) }\quad &(\alpha+\beta)\left(\alpha^{2}-\alpha \beta+\beta^{2}\right) \\\\ &=\alpha^{3}-\alpha^{2} \beta+\alpha \beta^{2}+\alpha^{2} \beta-\alpha \beta^{2}+\beta^{3} \\\\ &=\alpha^{3}+\beta^{3} \end{aligned}$ The roots of the equation $2 x^{2}+6 x-7=0$ are $\alpha$ and $\beta$. $\begin{aligned} \therefore\quad 2 x^{2}+6 x-7 &=2(x-\alpha)(x-\beta) \\\\ &=2 x^{2}-2(\alpha+\beta)+2 \alpha \beta \end{aligned}$ $\therefore \quad a+\beta=-3$ and $\alpha \beta=-\dfrac{7}{2}$ $\begin{aligned} \therefore\quad\quad \alpha^{2}+\beta^{2} &=(\alpha+\beta)^{2}-2 \alpha \beta \\\\ &=9+7 \\\\ &=16 \\\\ \text{(b) }\quad \alpha^{3}+\beta^{3} &=(\alpha+\beta)\left(\alpha^{2}-\alpha \beta+\beta^{2}\right) \\\\ &=(-3)\left(16+\frac{7}{2}\right) \\\\ &=-\frac{117}{2} \end{aligned}$ $\begin{aligned} \text{(c) }\quad (a-\beta)^{2} &=a^{2}-2 \alpha \beta+\beta^{2} \\\\ &=16+7 \\\\ &=23 \\\\ \therefore \alpha-\beta &=\sqrt{23} \end{aligned}$ $\begin{aligned} \text{(d)}\quad \text{Since}(\alpha-\beta)^{3} &=\alpha^{3}-3 \alpha^{2} \beta+3 \alpha \beta^{2}-\beta^{3} \\\\ &=\alpha^{3}-\beta^{3}-3 \alpha \beta(\alpha-\beta) \\\\ \alpha^{3}-\beta^{3} &=(\alpha-\beta)^{3}+3 \alpha \beta(\alpha-\beta) \\\\ &=23 \sqrt{23}+3\left(-\frac{7}{2}\right) \sqrt{23} \\\\ &=\frac{25}{2} \sqrt{23} \end{aligned}$




11. $f(x)=x^{2}+(k-3) x+4$
    
    The roots of the equation $f(x)=0$ are $\alpha$ and $\beta$.
    1. Find, in terms of $k$, the value of $\alpha^{2}+\beta^{2}$.
    
    Given that $4\left(\alpha^{2}+\beta^{2}\right)=7 \alpha^{2} \beta^{2}$,
    
    
    2. without solving the equation $f(x)=0$, form a quadratic equation, with integer coefficients, which has roots $\dfrac{1}{\alpha^{2}}$ and $\dfrac{1}{\beta^{2}}$
    3. find the possible values of $k$.



$f(x)=x^{2}+(k-3) x+4$ The roots of $f(x)=0$ are $\alpha$ and $\beta$. $\begin{aligned} &\therefore x^{2}+(k-3) x+4=(x-\alpha)(x-\beta) \\\\ &x^{2}+(k-3) x+4=x^{2}-(\alpha+\beta)+\alpha \beta \\\\ &\therefore-(\alpha+\beta)=k-3 \\\\ &\alpha+\beta=3-k \\\\ &\alpha \beta=4 \end{aligned}$ $\begin{aligned} \text { (a) }\quad\quad \alpha^{2}+\beta^{2} &=(\alpha+\beta)^{2}-2 \alpha \beta \\\\ &=(3-k)^{2}-8 \\\\ &=k^{2}-6 k+1 \\\\ \alpha\left(\alpha^{2}+\beta^{2}\right) &=7 \alpha^{2} \beta^{2} \\\\ \dfrac{\alpha^{2}+\beta^{2}}{d^{2} \beta^{2}} &=\dfrac{7}{4} \\\\ \dfrac{1}{\alpha^{2}}+\dfrac{1}{\beta^{2}} &=\dfrac{7}{4} \end{aligned}$ $\text { (b) }$ The quadratic equation Which has roots $\dfrac{1}{\alpha^{2}}$ and $\dfrac{1}{\beta^{2}}$ is $\begin{aligned} &\left(x-\dfrac{1}{\alpha^{2}}\right)\left(x-\dfrac{1}{\beta^{2}}\right)=0 \\\\ &x^{2}-\left(\dfrac{1}{\alpha^{2}}+\dfrac{1}{\beta^{2}}\right) x+\dfrac{1}{(\alpha \beta)^{2}}=0 \\\\ &x^{2}-\dfrac{7}{4} x+\dfrac{1}{16}=0 \\\\ &16 x^{2}-28+1=0\\\\ \text { (c) } &\alpha\left(\alpha^{2}+\beta^{2}\right)=7 \alpha^{2} \beta^{2} \\\\ & 4\left(k^{2}-6 k+1\right)=7(4)^{2} \\\\ & k^{2}-6 k+1=28 \\\\ & k^{2}-6 k-27=0 \\\\ & (k+3)(k-9)=0 \\\\ & k=-3 \text { or } k=9 \end{aligned}$




12. $f(x)=x^{2}+p x+7 \quad p \in \mathbb{R}$.
    
    The roots of the equation $\mathrm{f}(x)=0$ are $\alpha$ and $\beta$.
    1. Find, in terms of $p$ where necessary,
       1. $\alpha^{2}+\beta^{2}$,
       2. $\alpha^{2} \beta^{2}$.
    
    Given that $7\left(\alpha^{2}+\beta^{2}\right)=5 \alpha^{2} \beta^{2}$.
    
    
    2. find the possible values of $p$.
    
    Using the positive value of $p$ found in part (b) and without solving the equation $f(x)=0$,
    
    
    3. form a quadratic equation with roots $\dfrac{2 p}{\alpha^{2}}$ and $\dfrac{2 p}{\beta^{2}}$.



$f(x)=x^{2}+p x+7, p \in R$ The roots of $f(x)=0$ are $\alpha$ and $\beta$. $\begin{aligned} \therefore\quad x^{2}+p x+7 &=(x-\alpha)(x-\beta) \\\\ x^{2}+p x+7 &=x^{2}-(\alpha+\beta) x+\alpha \beta\\\\ \therefore\quad \alpha+\beta=-p\ & \text{ and }\ \alpha \beta=7\\\\ \text { (i) } \alpha^{2}+\beta^{2} &=(\alpha+\beta)^{2}-2 \alpha \beta \\\\ &=p^{2}-14\\\\ \text { (ii) }\alpha^{2} \beta^{2} &=(\alpha \beta)^{2}=49\\\\ 7\left(a^{2}+\beta^{2}\right) &=5 \alpha^{2} \beta^{2} \\\\ 7\left(p^{2}-14\right) &=5(49) \\\\ p^{2}-14 &=35 \\\\ p^{2} &=49 \\\\ p &=\pm 7 \end{aligned}$ The quadratic equations which has roots $\dfrac{2 p}{\alpha^{2}}$ and $\dfrac{2 p}{\beta^{2}}$ where $p>0$ is $\left(x-\dfrac{2 p}{\alpha^{2}}\right)\left(x-\dfrac{2 p}{\beta^{2}}\right)=0$ $\begin{aligned} &x^{2}-\left(\dfrac{2 p}{\alpha^{2}}+\dfrac{2 p}{\beta^{2}}\right) x+\dfrac{4 p^{2}}{\alpha^{2} \beta^{2}}=0 \\\\ &x^{2}-2 p\left(\dfrac{\alpha^{2}+\beta^{2}}{\alpha^{2} \beta^{2}}\right) x+\dfrac{4 p^{2}}{\alpha^{2} \beta^{2}}=0 \\\\ &x^{2}-14\left(\dfrac{49-14}{49}\right) x+\dfrac{4(49)}{49}=0 \\\\ &x^{2}-10 x+4=0 \end{aligned}$




13. The equation $3 x^{2}-5 x+4=0$ has roots $\alpha$ and $\beta$. Without solving this equation, form a quadratic equation with integer coefficients that has roots $\alpha+\dfrac{1}{2 \beta}$ and $\beta+\dfrac{1}{2 \alpha}$.



The equation $3 x^{2}-5 x+4=0$ has roots $\alpha$ and $\beta$. $\therefore \quad 3 x^{2}-5 x+4=3(x-\alpha)(x-\beta)$ $\quad\quad 3 x^{2}-5 x+4=3 x^{2}-3(\alpha+\beta) x+3 \alpha \beta$ $\therefore \quad \alpha+\beta=\dfrac{5}{3}$ and $\alpha \beta=\dfrac{4}{3}$ The quadratic equation which has roots $\alpha+\dfrac{1}{2 \beta}$ and $\beta+\dfrac{1}{2 \alpha}$ is $\begin{aligned} &{\left[x-\left(\alpha+\dfrac{1}{2 \beta}\right)\right]\left[x-\left(\beta+\dfrac{1}{2 \alpha}\right)\right]=0} \\\\ &x^{2}-\left(\alpha+\beta+\dfrac{1}{2 \alpha}+\dfrac{1}{2 \beta}\right) x+\left(\alpha \beta+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{4 \alpha \beta}\right)=0 \end{aligned}$ $x^{2}-\left(\alpha+\beta+\dfrac{\alpha+\beta}{2 \alpha \beta}\right) x+\left(1+\alpha \beta+\dfrac{1}{4 \alpha \beta}\right)=0$ $x^{2}-\left(\dfrac{5}{3}+\dfrac{5 / 3}{8 / 3}\right) x+\left(1+\dfrac{4}{3}+\dfrac{1}{16 / 3}\right)=0$ $x^{2}-\dfrac{5}{24} x+\dfrac{121}{48}=0$ $48 x^{2}-110 x+121=0$




14. The roots of the equation $x^{2}+3 x-5=0$ are $\alpha$ and $\beta$.
    1. Without solving the equation, find
       1. the value of $\alpha^{2}+\beta^{2}$.
       2. the value of $\alpha^{4}+\beta^{4}$.
    
    Given that $\alpha>\beta$ and without solving the equation,
    
    
    2. show that $\alpha-\beta=\sqrt{29}$.
    3. Factorise $\alpha^{4}-\beta^{4}$ completely.
    4. Hence find the exact value of $\alpha^{4}-\beta^{4}$.
    
    Given that $\beta^{4}=p+q \sqrt{29}$ where $p$ and $q$ are positive constants,
    
    
    5. find the value of $p$ and the value of $q$.



The roots of the equation $x^{2}+3 x-5=0$ are $\alpha$ and $\beta$. $\begin{aligned} \therefore \quad x^{2}+3 x-5&=(x-\alpha)(x-\beta) \\\\ x^{2}+3 x-5 & =x^{2}-(\alpha+\beta) x+\alpha \beta \\\\ \therefore\quad \alpha +\beta=-3 & \text { and } \alpha \beta=-5 \end{aligned}$ $\begin{aligned} \text { (a) (i) }\qquad \alpha^{2}+\beta^{2} &=(\alpha+\beta)^{2}-2 d \beta \\\\ &=9-2(-5) \\\\ &=19\\\\ \quad \text {(ii) }\qquad \alpha^{4}+\beta^{4} &=\left(\alpha^{2}\right)^{2}+\left(\beta^{2}\right)^{2} \\\\ &=\left(\alpha^{2}+\beta^{2}\right)^{2}-2 \alpha^{2} \beta^{2} \\\\ &=19^{2}-2(-5)^{2} \\\\ &=311\\\\ \text { (b) }\qquad\quad (\alpha-\beta)^{2} &=\alpha^{2}+\beta^{2}-2 \alpha \beta \\\\ &=19-2(-5) \\\\ &=29 \\\\ \therefore\quad \alpha-\beta &=\sqrt{29}\\\\ \text { (c) }\qquad\quad \alpha^{4}-\beta^{4} &=\left(\alpha^{2}+\beta^{2}\right)\left(\alpha^{2}-\beta^{2}\right) \\\\ &=\left(\alpha^{2}+\beta^{2}\right)(\alpha+\beta)(\alpha-\beta) \\\\ \text { (d) }\qquad\quad \alpha^{4}-\beta^{4} &=(19)(-3)(\sqrt{29}) \\\\ &=-57 \sqrt{29} \end{aligned}$ $\begin{aligned} \text { (e) } \alpha^{4}+\beta^{4}-\left(\alpha^{4}-\beta^{4}\right)&=311+57 \sqrt{29} \\\\ 2 \beta^{4}&=311+57 \sqrt{29} \\\\ \beta^{4}&=\dfrac{311}{2}+\dfrac{57}{2} \sqrt{29}\\\\ \beta^{4}&=p+q \sqrt{29}\ \text{( given )}\\\\ \therefore\quad p=\dfrac{311}{2},\ & q=\dfrac{57}{2} \end{aligned}$




15. It is given that $\alpha$ and $\beta$ are such that $a+\beta=-\dfrac{5}{2}$ and $\alpha \beta=-5$.
    1. Form a quadratic equation with integer coefficients that has roots $\alpha$ and $\beta$
    
    Without solving the equation found in part (a),
    
    
    2. find the value of
       1. $\alpha^{2}+\beta^{2}$.
       2. $a^{3}+\beta^{3}$.
    3. Hence form a quadratic equation with integer coefficients that has roots $\left(\alpha-\dfrac{1}{\alpha^{2}}\right)$ and $\left(\beta-\dfrac{1}{\beta^{2}}\right)$.
    
    
    
    

    $\alpha+\beta=-\dfrac{5}{2},\ \alpha \beta=-5$ (a) The quadratic equation which has roots $\alpha$ and $\beta$ is $\begin{aligned} &(x-\alpha)(x-\beta)=0 \\\\ &x^{2}-(\alpha+\beta) x+\alpha \beta=0 \\\\ &x^{2}-\left(-\dfrac{5}{2}\right) x-5=0 \\\\ &2 x^{2}+5 x-10=0 \end{aligned}$ $\begin{aligned} \text {(b) (i) } \alpha^{2}+\beta^{2} &=(\alpha+\beta)^{2}-2 \alpha \beta \\\\ &=\dfrac{25}{4}-2(-5)=\dfrac{65}{4}\\\\ \quad \text { (ii) } (\alpha+\beta)^{3} &=\alpha^{3}+3 \alpha^{2} \beta+3 \alpha \beta^{2}+\beta^{3} \\\\ &=\alpha^{3}+\beta^{3}+3 \alpha \beta(\alpha+\beta) \\\\ \therefore\quad \alpha^{3}+\beta^{3} &=(\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta) \\\\ &=-\dfrac{125}{8}+15\left(-\dfrac{5}{2}\right) \\\\ &=-\dfrac{425}{8} \end{aligned}$ $\text {(c)}$ The quadratic equation which has roots $\alpha-\dfrac{1}{\alpha^{2}}$ and $\beta-\dfrac{1}{\beta^{2}}$ is $\begin{aligned} &{\left[x-\left(\alpha-\dfrac{1}{\alpha^{2}}\right)\right]\left[x-\left(\beta-\dfrac{1}{\beta^{2}}\right)\right]=0} \\\\ &x^{2}-\left(\alpha-\dfrac{1}{\alpha^{2}}+\beta-\dfrac{1}{\beta^{2}}\right) x+\left(\alpha \beta-\dfrac{\alpha}{\beta^{2}}-\dfrac{\beta}{\alpha^{2}}+\dfrac{1}{\alpha^{2} \beta^{2}}\right)=0 \\\\ &x^{2}-\left(\alpha+\beta-\dfrac{\alpha^{2}+\beta^{2}}{\alpha^{2} \beta^{2}}\right) x+\left(\alpha \beta-\dfrac{\alpha^{3}+\beta^{3}}{d^{2} \beta^{2}}+\dfrac{1}{\alpha^{2} \beta^{2}}\right)=0 \\\\ &x^{2}-\left(-\dfrac{5}{2}-\dfrac{65 / 4}{25}\right) x+\left(-5-\dfrac{-425 / 8}{25}+\dfrac{1}{25}\right)=0 \\\\ &200 x^{2}+630 x-567=0 \end{aligned}$

    
    
    

Factorial Expression : Exercise

Factorials

We define $\boldsymbol{n} !=\boldsymbol{n}(\boldsymbol{n}-\mathbf{1})(\boldsymbol{n}-\mathbf{2}) \cdots \mathbf{3} \cdot \mathbf{2} \cdot \mathbf{1}$ if $n$ is a nonnegative integer. An empty product is normally defined to be 1 . With this convention, $0 !=1$ An alternative is to define $\boldsymbol{n} !$ recursively on the nonnegative integers. $\boldsymbol{n} != \begin{cases}1 & \text { if } \boldsymbol{n}=\mathbf{0} \\ \boldsymbol{n}(\boldsymbol{n}-\mathbf{1}) ! & \text { if } \boldsymbol{n} \geq \mathbf{1}\end{cases}$

Exercise

1. Evaluate
   $\begin{array}{lll} \text{(a)}\ 2 !& \text{(b)}\ 3 !& \text{(c)}\ 4 !\\\\ \text{(e)}\ 5 !& \text{(f)}\ 6 !& \text{(g)}\ 10 ! \end{array}$





$\begin{aligned} \text{(a)}\ &2 !=2 \times 1=2 \\\\ \text{(b)}\ &3 !=3 \times 2 \times 1=6 \\\\ \text{(c)}\ &4 !=4 \times 3 \times 2 \times 1=24 \\\\ \text{(d)}\ &5 !=5 \times 4 \times 3 \times 2 \times 1=120 \\\\ \text{(e)}\ &6 !=6 \times 5 \times 4 \times 3 \times 2 \times 1=720 \\\\ \text{(f)}\ &10 !=10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1=362880 \end{aligned}$



2. Express in factorial form:
   $\begin{array}{l} \text{(a)}\ 4 \times 3 \times 2 \times 1 \\\\ \text{(b)}\ 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \quad \\\\ \text{(c)}\ 6 \times 5 \\\\ \text{(d)}\ 8 \times 7 \times 6\\\\ \text{(e)}\ 10 \times 9 \times 8 \times 7 \\\\ \text{(f)}\ 15 \times 14 \times 13 \times 12\\\\ \text{(g)}\ \dfrac{9 \times 8 \times 7}{3 \times 2 \times 1} \\\\ \text{(h)}\ \dfrac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1}\\\\ \text{(i)}\ \dfrac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} \end{array}$





$\begin{aligned} \text{(a)}\ &\quad 4 \times 3 \times 2 \times 1\\\\ &=4 !\\\\ \text{(b)}\ &\quad 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\\\\ &=7 !\\\\ \text{(c)}\ &\quad 6 \times 5\\\\ &=\dfrac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1}\\\\ &=\dfrac{6 !}{4 !}\\\\ \text{(d)}\ &\quad 8 \times 7 \times 6\\\\ &=\dfrac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{5 \times 4 \times 3 \times 2 \times 1}\\\\ &=\dfrac{8 !}{5 !}\\\\ \text{(e)}\ &\quad 10 \times 9 \times 8 \times 7\\\\ &=\dfrac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1}\\\\ &=\dfrac{10 !}{6 !}\\\\ \text{(f)}\ &\quad 15 \times 14 \times 13 \times 12\\\\ &=\dfrac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}\\\\ &=\dfrac{15 !}{11 !}\\\\ \text{(g)}\ &\quad \dfrac{9 \times 8 \times 7}{3 \times 2 \times 1}\\\\ &=\dfrac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(6 \times 5 \times 4 \times 3 \times 2 \times 1)}\\\\ &=\dfrac{9 !}{3 ! 6 !}\\\\ \text{(h)}\ &\quad \dfrac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1}\\\\\ &=\dfrac{13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(4 \times 3 \times 2 \times 1)(9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}\\\\ &=\dfrac{13 !}{4 ! 9 !}\\\\ \text{(i)}\ &\quad \dfrac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1}\\\\ &=\dfrac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(5 \times 4 \times 3 \times 2 \times 1)( 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}\\\\ &=\dfrac{15 !}{5 ! 10 !} \end{aligned}$ $\begin{aligned} &\textbf{Alternative Method }\\\\ \text{(a)}\ &\quad 4 \times 3 \times 2 \times 1\\\\ &=4 !\\\\ \text{(b)}\ &\quad 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\\\\ &=7 !\\\\ \text{(c)}\ &\quad 6 \times 5=\dfrac{6 \times 5 \times 4 !}{4 !}\\\\ &=\dfrac{6 !}{4 !}\\\\ \text{(d)}\ &\quad 8 \times 7 \times 6\\\\ &=\dfrac{8 \times 7 \times 6 \times 5 !}{5 !}\\\\ &=\dfrac{8 !}{5 !}\\\\ \text{(e)}\ &\quad 10 \times 9 \times 8 \times 7\\\\ &=\dfrac{10 \times 9 \times 8 \times 7 \times 6 !}{6 !}\\\\ &=\dfrac{10 !}{6 !}\\\\ \text{(f)}\ &\quad 15 \times 14 \times 13 \times 12=\dfrac{15 \times 14 \times 13 \times 12 \times 11 !}{11 !}\\\\ &=\dfrac{15 !}{11 !}\\\\ \text{(g)}\ &\quad \dfrac{9 \times 8 \times 7}{3 \times 2 \times 1}\\\\ &=\dfrac{9 \times 8 \times 7 \times 6 !}{3 \times 2 \times 1 \times 6 !}\\\\ &=\dfrac{9 !}{3 ! 6 !}\\\\ \text{(h)}\ &\quad \dfrac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1}\\\\ &=\dfrac{13 \times 12 \times 11 \times 10 \times 9 !}{4 \times 3 \times 2 \times 1 \times 9 !}\\\\ &=\dfrac{13 !}{4 ! 9 !} \\\\ \text{(i)}\ &\quad \dfrac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1}\\\\ &=\dfrac{15 \times 14 \times 13 \times 12 \times 11 \times 10 !}{5 \times 4 \times 3 \times 2 \times 1 \times 10 !}\\\\ &=\dfrac{15 !}{5 ! 10 !} \end{aligned}$



3. Simplify without using a calculator:
   $\begin{array}{ll} \text{(a)}\ \dfrac{7 !}{6 !}& \text{(b)}\ \dfrac{8 !}{6 !}\\\\ \text{(c)}\ \dfrac{12 !}{10 !}& \text{(d)}\ \dfrac{120 !}{119 !}\\\\ \text{(e)}\ \dfrac{10 !}{8 ! \times 2 !}& \text{(f)}\ \dfrac{100 !}{98 ! \times 2 !}\\\\ \text{(g)}\ \dfrac{7 !}{3 !}& \text{(h)}\ \dfrac{8 !}{5 !}\\\\ \text{(i)}\ \dfrac{4 !}{2 ! 2 !}& \text{(j)}\ \dfrac{6 !}{3 ! 2 !}\\\\ \text{(k)}\ \dfrac{6 !}{(3 !)^{2}}& \text{(l)}\ \dfrac{5 !}{3 !} \times \dfrac{7 !}{4 !} \end{array}$





$\begin{array}{l} \text{(a)}\ \dfrac{7 !}{6 !}=\dfrac{7 \times 6 !}{6 !}=7\\\\ \text{(b)}\ \dfrac{8 !}{6 !}=\dfrac{8 \times 7 \times 6 !}{6 !}=56\\\\ \text{(c)}\ \dfrac{12 !}{10 !}=\dfrac{12 \times 11 \times 10 !}{10 !}=132\\\\ \text{(d)}\ \dfrac{120 !}{119 !}=\dfrac{120 \times 119 !}{119 !}=120\\\\ \text{(e)}\ \dfrac{10 !}{8 ! \times 2 !}=\dfrac{10 \times 9 \times 8 !}{8 ! \times(2 \times 1)}=45\\\\ \text{(f)}\ \dfrac{100 !}{98 ! \times 2 !}=\dfrac{100 \times 99 \times 98 !}{98 ! \times 2 \times 1}=4950\\\\ \text{(g)}\ \dfrac{7 !}{3 !}=\dfrac{7 \times 6 \times 5 \times 4 \times 3 !}{3 !}=840\\\\ \text{(h)}\ \dfrac{8 !}{5 !}=\dfrac{8 \times 7 \times 6 \times 5 !}{5 !}=336\\\\ \text{(i)}\ \dfrac{4 !}{2 ! 2 !}=\dfrac{4 \times 3 \times 2 !}{2 !(2 \times 1)}=6\\\\ \text{(j)}\ \dfrac{6 !}{3 ! 2 !}=\dfrac{6 \times 5 \times 4 \times 3 !}{3 ! \times(2 \times 1)}=60\\\\ \text{(k)}\ \dfrac{6 !}{(3 !)^{2}}=\dfrac{6 \times 5 \times 4 \times 3 !}{3 !(3 \times 2 \times 1)}=20\\\\ \text{(l)}\ \dfrac{5 !}{3 !} \times \dfrac{7 !}{4 !}=(5 \times 4) \times(7 \times 6 \times 5)=4200 \end{array}$



4. Simplify:
   $\begin{array}{l} \text{(a)}\ \dfrac{n !}{(n-1) !}\\\\ \text{(b)}\ \dfrac{(n+2) !}{n !}\\\\ \text{(c)}\ \dfrac{(n+1) !}{(n-1) !} \end{array}$





$\begin{aligned} \text{(a)}\ &\dfrac{n !}{(n-1) !}\\\\\ &=\dfrac{n(n-1) !}{(n-1) !}\\\\ &=n \\\\ \text{(b)}\ &\dfrac{(n+2) !}{n !}\\\\ &=\dfrac{(n+2)(n+1) n !}{n !}\\\\ &=n^{2}+3 n+2 \\\\ \text{(c)}\ &\dfrac{(n+1) !}{(n-1) !}\\\\ &=\dfrac{(n+1) n(n-1) !}{(n-1) !}\\\\ &=n^{2}+n \end{aligned}$



5. Rewrite each of the following using factorial notation.
   $\begin{array}{l} \text{(a)}\ n(n-1)(n-2)(n-3)\\\\ \text{(b)}\ n(n-1)(n-2)(n-3)(n-4)(n-5)\\\\ \text{(c)}\ \dfrac{n(n-1)(n-2)}{5 \times 4 \times 3 \times 2 \times 1}\\\\ \text{(d)}\ \dfrac{n(n-1)(n-2)(n-3)(n-4)}{3 \times 2 \times 1} \end{array}$





$\begin{aligned} \text{(a)}\ &n(n-1)(n-2)(n-3)\\\\ &=\frac{n(n-1)(n-2)(n-3)(n-4) !}{(n-4) !}\\\\ &=\frac{n !}{(n-4) !} \\\\ \text{(b)}\ &n(n-1)(n-2)(n-3)(n-4)(n-5)\\\\ &=\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6) !}{(n-6) !}\\\\ &=\frac{n !}{(n-6) !} \\\\ \text{(c)}\ &\frac{n(n-1)(n-2)}{5 \times 4 \times 3 \times 2 \times 1}\\\\ &=\frac{n(n-1)(n-2)(n-3) !}{5 !(n-3) !}=\frac{n !}{5 !(n-3) !} \\\\ \text{(d)}\ &\frac{n(n-1)(n-2)(n-3)(n-4)}{3 \times 2 \times 1}\\\\ &=\frac{n(n-1)(n-2)(n-3)(n-4)(n-5) !}{3 !(n-5) !}\\\\ &=\frac{n !}{3 !(n-5) !} \end{aligned}$



6. Express the following as a single factorial notation.
   (a) $n !(n+1)$
   (b) $(n-1) !\left(n^{2}+n\right)$
   (c) $(n+4)(n+5)(n+3) !$
   (d) $n !\left(n^{2}+3 n+2\right)$
   (e) $(n+1)(n+2)(n+3)$
   (f) $(n-3)(n-4)(n-5)$





$\begin{aligned} \text { (a) } & n !(n+1) \\\\ &=(n+1) n ! \\\\ &=(n+1) ! \\\\ \text { (b) } &(n-1) !\left(n^{2}+n\right) \\\\ &=\left(n^{2}+n\right)(n-1) ! \\\\ &=(n+1) n(n-1) ! \\\\ &=(n+1) ! \\\\ &=(n+5)(n+4)(n+3) ! \\\\ &=(n+5) !\\\\ \text { (c) } &(n+4)(n+5)(n+3) ! \\\\ &=(n+5)(n+4)(n+3) ! \\\\ &=(n+5) !\\\\ \text { (d) } & n !\left(n^{2}+3 n+2\right) \\\\ =&(n+2)(n+1) n ! \\\\ =&(n+2) !\\\\ \text { (e) } &(n+1)(n+2)(n+3) \\\\ =& \frac{(n+3)(n+2)(n+1) n !}{n !} \\\\ =& \frac{(n+3) !}{n !}\\\\ \text { (f) } & (n-3)(n-4)(n-5) \\\\ &=\frac{(n-3)(n-4)(n-5)(n-6) !}{(n-6) !} \\\\ &=\frac{(n-3) !}{(n-6) !} \end{aligned}$



7. Write as a product by factorizing:
   (a) $5 !+4 !$
   (b) $11 !-10 !$
   (c) $5 !+7 !$
   (d) $12 !-10 !$
   (e) $9 !+8 !+7 !$
   (f) $7 !-6 !+8 !$
   (g) $12 !-2 \times 11 !$
   (h) $3 \times 9 !+5 \times 8 !$





$\begin{aligned} \text { (a) } & \quad 5 !+4 ! \\\\ &= 5 \times 4 !+4 ! \\\\ &=(5+1) 4 ! \\\\ &= 6 \times 4 ! \\\\ \text { (b) } & \quad 11 !-10 ! \\\\ &=(11-1) 10 ! \\\\ &= 10 \times 10 !\\\\ \text { (c) } & \quad 5 !+7 ! \\\\ &= 5 !+7 \times 6 \times 5 ! \\\\ &=(1+42) 5 ! \\\\ &= 43 \times 5 ! \\\\ \text { (d) } & \quad 12 !-10 ! \\\\ &= 12 \times 11 \times 10 !-10 ! \\\\ &=(132-1) 10 ! \\\\ &= 131 \times 10 !\\\\ \text { (e) } & \quad 9 !+8 !+7 ! \\\\ &= 9 \times 8 \times 7 !+8 \times 7 !+7 ! \\\\ &=(72+8+1) 7 ! \\\\ &= 81 \times 7 ! \\\\ \text { (f) } & \quad 7 !-6 !+8 ! \\\\ &= 7 \times 6 !-6 !+8 \times 7 \times 6 ! \\\\ &=(7-1+56) 6 ! \\\\ &= 62 \times 6 !\\\\ \text { (g) } & \quad 12 !-2 \times 11 ! \\\\ &= 12 \times 11 !-2 \times 11 ! \\\\ &=(12-2) 11 ! \\\\ &= 10 \times 11 ! \\\\ \text { (h) } & \quad 3 \times 9 !+5 \times 8 ! \\\\ &= 3 \times 9 \times 8 !+5 \times 8 ! \\\\ &=(27+5) 8 ! \\\\ &= 32 \times 8 ! \end{aligned}$



8. Simplify by factorizing:
   $\begin{array}{l} \text{(a)}\ \dfrac{12 !-11 !}{11}\\\\ \text{(b)}\ \dfrac{10 !+9 !}{11}\\\\ \text{(c)}\ \dfrac{10 !-8 !}{89}\\\\ \text{(d)}\ \dfrac{10 !-9 !}{9 !}\\\\ \text{(e)}\ \dfrac{6 !+5 !-4 !}{4 !}\\\\ \text{(f)}\ \dfrac{n !+(n-1) !}{(n-1) !}\\\\ \text{(g)}\ \dfrac{n !-(n-1) !}{n-1}\\\\ \text{(h)}\ \dfrac{(n+2) !+(n+1) !}{n+3} \end{array}$





$\begin{aligned} \text { (a) } & \quad \dfrac{12 !-11 !}{11} \\\\ &= \dfrac{12 \times 11 !-11 !}{11} \\\\ &= \dfrac{(12-1) 11 !}{11} \\\\ &= \dfrac{11 \times 11 !}{11} \\\\ &= 11 !\\\\ \text { (b) } & \quad \dfrac{10 !+9 !}{11} \\\\ &= \dfrac{10 \times 9 !+9 !}{11} \\\\ &= \dfrac{(10+1) 9 !}{11} \\\\ &= \dfrac{11 \times 9 !}{11} \\\\ &= 9 !\\\\ \text { (c) } & \quad \dfrac{10 !-8 !}{89} \\\\ &= \dfrac{10 \times 9 \times 8 !-8 !}{89} \\\\ &= \dfrac{(90-1) \times 8 !}{89} \\\\ &= \dfrac{89 \times 8 !}{89} \\\\ &= 8 !\\\\ \text { (d) } & \quad \dfrac{10 !-9 !}{9} \\\\ &= \dfrac{10 \times 9 !-9 !}{9} \\\\ &= \dfrac{(10-1) 9 !}{9} \\\\ &= \dfrac{9 \times 9 !}{9} \\\\ &= 9 !\\\\ \text { (e) } & \quad \dfrac{6 !+5 !-4 !}{4 !} \\\\ &= \dfrac{6 \times 5 \times 4 !+5 \times 4 !-4 !}{4 !} \\\\ &= \dfrac{(30+5-1) 4 !}{4 !} \\\\ &= 34\\\\ \text { (f) } & \quad \dfrac{n !+(n-1) !}{(n-1) !} \\\\ &=\dfrac{n(n-1) !+(n-1) !}{(n-1) !} \\\\ &=\dfrac{(n+1)(n-1) !}{(n-1) !} \\\\ &=n+1\\\\ \text { (g) } & \quad \dfrac{n !-(n-1) !}{n-1} \\\\ &=\dfrac{n(n-1) !-(n-1) !}{n-1} \\\\ &=\dfrac{(n-1)(n-1) !}{n-1} \\\\ &=(n-1) !\\\\ \text { (h) } & \quad \dfrac{(n+2) !+(n+1) !}{n+3} \\\\ &= \dfrac{(n+2)(n+1) !+(n+1) !}{n+3} \\\\ &= \dfrac{(n+2+1)(n+1) !}{n+3} \\\\ &= \dfrac{(n+3)(n+1) !}{n+3} \\\\ &=(n+1) ! \end{aligned}$

Arithmetic Progression : Problems and Solutions - Part (2)

Post တစ်ပုဒ်ထဲမှာ math loading (rendering) ကြာနေသောကြောင့် နှစ်ပိုင်း ခွဲလိုက်ရပါတယ်။

1. If $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ term of an A.P. are $a, b, c$ respectively, then show that $(a-b) r$ $+(b-c) p$ $+(c-a) q=0$.





Let the first term and the comnon difference of given A.P. be $A$ and $D$. By the proldem, $u_{p}=a$ $A+(p-1) D=a$ $u_{q}=b$ $A+(q-1) D=b$ $u_{r}=c$ $A+(r-1) D=c$ $\therefore\ (a-b) r=(p-q) D r$ $\hspace{2.2cm}=(p r-q r) D \ldots(1)$ $\quad\ (p-c) p=(q-r) D p$ $\hspace{2.2cm} =(p q-p r) D \ldots(2)$ $\quad\ (c-a) q =(r-p) D q$ $\hspace{2.2cm} =(q r-p q) D \ldots(3)$ Summing equations $(1),(2)$ and $(3)$ $(a-b) r+(b-c) p+(c-a) q=0$



2. Show that the sum of $(m+n)^{\text {th }}$ and $(m-n)^{\text {th }}$ term of an A.P is equal to twice the $m^{\text {th }}$ term.





Let the first tern be $a$ and the common difference be $d$ for the given A.P. $\therefore\ u_{m+n}=a+(m+n-1) d$ $\quad\ u_{m-n}=a+(m-n-1) d$ $\quad\ u_{m+n}+u_{m-n}=2 a+(2 n-1) d$ $\hspace{3.2cm} =2[a+(m-1) d]$ $\hspace{3.2cm} =2 u_{m}$



3. If $(m+1)^{\text {th }}$ term of an AP is twice the $(n+1)^{\text {th }}$ term, prove that $(3 m+1)^{\text {th }}$ term is twice the $(m+n+1)^{\text {th }}$ term.





Let the first tern be $a$ and the common difference be $d$ for the given A.P. By the problem, $u_{m+1}=2 u_{n+1}$ $a+m d=2(a+n d)$ $a+m d=2 a+2 n d$ $a=m d-2 n d$ $u_{m+n+1} =a+(m+n) d$ $\hspace{1.5cm}=m d-2 n d+m d+n d$ $\hspace{1.5cm}=2 m d-n d$ $\displaystyle u_{3 m+1} =a+3 m d$ $\hspace{1.5cm}=m d-2 n d+3 m d$ $\hspace{1.5cm}=4 m d-2 n d$ $\hspace{1.5cm}=2(2 m d-n d)$ $\hspace{1.5cm}=2 u_{m+n+1}$



4. The digits of a positive integer having three digits are in A.P. The sum of the digits is 15 and the number obtained by reversing the digits is 594 less than the original number. Find the number.





Let the hundredis digit, ten's digit and one's digit of a positive integer be $a, b$ and $c$ respectively. By the problem, $a, b, c$ are in A.P. $\therefore\ a=a$ $\quad\ b=a+d$ $\quad\ c=a+2 d$ $\quad\ a+b+c=15$ (given) $\quad\ 3 a+3 d=15$ $\therefore\ a+d=5\Rightarrow b=5$ $\therefore$ given integer $=100 a+10 b+c$ Original neumber - New formed nunber $=594$ $100 a+10 b+c-(100 c+10 b+a)=594$ $99 a-99 c=594$ $\quad\ a-c=6$ $\quad -2 d=6$ $\quad\quad d=-3$ $\therefore\ a-3=5$ $\quad\ a=8$ $\therefore\ c=2$ $\therefore$ The number is $852 .$



5. If $\dfrac{b+c-a}{a}, \dfrac{c+a-b}{b}, \dfrac{a+b-c}{c}$ are in A.P., then prove that $\dfrac{1}{a}$, $\dfrac{1}{b}$, $\dfrac{1}{c}$ are in A.P.





$\dfrac{b+c-a}{a}, \dfrac{c+a-b}{b}, \dfrac{a+b-c}{c}$ are in A.P. $\dfrac{c+a-b}{b}-\dfrac{b+c-a}{a}=\dfrac{a+b-c}{c}-\dfrac{c+a-b}{b}$ $\dfrac{c+a}{b}-1-\dfrac{b+c}{a}+1=\dfrac{a+b}{c}-1-\dfrac{c+a}{b}+1$ $\dfrac{c+a}{b}-\dfrac{b+c}{a}=\dfrac{a+b}{c}-\dfrac{c+a}{b}$ $\dfrac{a^{2}+a c-b^{2}-b c}{a b}=\dfrac{b^{2}+a b-c^{2}-a c}{b c}$ $\dfrac{a^{2}-b^{2}+a c-b c}{a}=\dfrac{b^{2}-c^{2}+a b-a c}{c}$ $\dfrac{(a-b)(a+b)+c(a-b)}{a}=\dfrac{(b-c)(b+c)+a(b-c)}{c}$ $\dfrac{(a-b)(a+b+c)}{a}=\dfrac{(b-c)(a+b+c)}{c}$ $\dfrac{a-b}{a}=\dfrac{b-c}{c}$ $\therefore\ a c-b c=a b-a c$ $\therefore\ \dfrac{a c}{a b c}-\dfrac{b c}{a b c}=\dfrac{a b}{a b c}-\dfrac{a c}{a b c}$ $\quad\ \dfrac{1}{b}-\dfrac{1}{a}=\dfrac{1}{c}-\dfrac{1}{b}$ $\therefore\ \dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}$ is an A.P.



6. If $a, b, c$ are in A.P., then prove that $(a-c)^{2}=4\left(b^{2}-a c\right)$.





$\quad\ a, b, c$ are in A.P. $\therefore\ b-a=c-b$ $\quad\ a+c=2 b$ $\quad\ a+c-2c=2 b-2 c$ $\quad\ a-c=2(b-c)$ $\quad\ (a-c)^{2}=4(b-c)^{2}$ $\quad\ (a-c)^{2} =4\left(b^{2}-2 b c+c^{2}\right) $ $\hspace{2.15cm}=4\left(b^{2}-(a+c) c+c^{2}\right) $ $\hspace{2.15cm}=4\left(b^{2}-a c-c^{2}+c^{2}\right) $ $\hspace{2.15cm}=4\left(b^{2}-a c\right)$



7. If $a, b, c$ are in A.P., then prove that $b+c, c+a, a+b$ are also in A.P.





$\quad\ a, b, c$ are in A.P. $\therefore \ b-a=c-b$ $\quad\ 2 b=c+a$ $\quad\ 2 b+c+a=c+a+c+a$ $\quad\ (b+c)+(a+b)=(c+a)+(c+a)$ $\therefore\ (c+a)-(b+c)=(a+b)-(c+a)$ $\therefore\ b+c, c+a, a+b$ are in A.P.



8. If $a, b, c$ are in A.P., then prove that $\dfrac{1}{b c}$, $\dfrac{1}{c a}$, $\dfrac{1}{a b}$ are also in A.P.





$\begin{aligned} & a, b, c \text{ are in A.P.}\\\\ \therefore\ &b-a =c-b\\\\ &\dfrac{b}{a b c}-\dfrac{a}{a b c}=\dfrac{c}{a b c}-\dfrac{b}{a b c}\\\\ &\dfrac{1}{a c}-\dfrac{1}{b c} =\dfrac{1}{a b}-\dfrac{1}{a c}\\\\ &\dfrac{1}{b c},\ \dfrac{1}{c a},\ \dfrac{1}{a b}\ \text{ are in A.P.} \end{aligned}$



9. If $a, b, c$ are in A.P., then prove that $(b+c-a)$,$(c+a-b)$,$(a+b-c)$ are in AP.





$\begin{aligned} &a, b, c \text { are in A.P.} \\\\ \therefore\ &b-a=c-b \\\\ &a-b=b-c \\\\ &2 a-2 b=2 b-2 c \\\\ &c+2 a-2 b-c=a+2 b-2 c-a \\\\ &c+a-b-b-c+a=a+b-c-c-a+b \\\\ &(c+a-b)-(b+c-a)=(a+b-c)-(c+a-b) \\\\ \therefore\ &b+c-a),\ (c+a-b),\ (a+b-c)\ \text { are in A.P.} \end{aligned}$



10. If $a, b, c$ are in A.P., then prove that $a^{2}(b+c)$, $b^{2}(c+a)$, $c^{2}(a+b)$ are also in A.P.





$\begin{aligned} &a, b, c \text { are in A.P.} \\\\ \therefore\ &b-a=c-b \\\\ \therefore\ &a+c=2 b \\\\ &a^{2}(b+c)+c^{2}(a+b) \\\\ =& a^{2} b+a^{2} c+c^{2} a+c^{2} b \\\\ =& a^{2} b+c a(a+c)+c^{2} b \\\\ =& a b+c a(2 b)+c^{2} b \\\\ =& a^{2} b+2 a b c+c b \\\\ =& a^{2} b +a b c+a b c+c^{2} b\\\\ =& a b(a+c)+b c(a+c) \\\\ =& a b(2 b)+b c(2 b) \\\\ =& 2 a b^{2}+2 b^{2} c \\\\ =& 2 b^{2}(c+a)\\\\ \therefore\ & 2 b^{2}(c+a)=a^{2}(b+c)+c^{2}(a+b) \\\\ \therefore\ & b^{2}(c+a)+b^{2}(c+a)=a^{2}(b+c)+c^{2}(a+b) \\\\ \therefore\ & b^{2}(c+a)-a^{2}(b+c)=c^{2}(a+b)-b^{2}(c+a) \\\\ \therefore\ & a^{2}(b+c), b^{2}(c+a), c^{2}(a+b) \text { are in A.P.} \end{aligned}$



11. If $a, b, c$ are in A.P., then prove that $b c-a^{2}$, $c a-b^{2}$, $a b-c^{2}$ are in AP.





$\begin{aligned} &a, b, c \text { are in A.P.} \\\\ \therefore\ &b-a=c-b \\\\ \therefore\ &(b-a)(a+b+c)=(c-b)(a+b+c) \\\\ &(b-a)(b+a)+(b-a) c=(c-b)(c+b)+(c-b) a \\\\ &b^{2}-a^{2}+b c-c a=c^{2}-b^{2}+c a-a b \\\\ &\left(b c-a^{2}\right)-\left(c a-b^{2}\right)=\left(c a-b^{2}\right)-\left(a b-c^{2}\right) \\\\ &\text{Multiply both sides with}\ -1,\\\\ &\left(c a-b\right)^{2}-\left(b c-a^{2}\right)=\left(a b-c^{2}\right)-\left(c a-b^{2}\right) \\\\ \therefore\ &b c-a^{2}, c a-b^{2}, a b-c^{2} \text { are in A.P.} \end{aligned}$



12. If $a, b, c$ are in A.P., then prove that $\dfrac{1}{\sqrt{b}+\sqrt{c}}$, $\dfrac{1}{\sqrt{c}+\sqrt{a}}$, $\dfrac{1}{\sqrt{a}+\sqrt{b}}$ are also in A.P.





$\begin{aligned} &a,\ b,\ c \text { are in A.P.} \\\\ \therefore\ & b-a=c-b\\\\ &(\sqrt{b})^{2}-(\sqrt{a})^{2}=(\sqrt{e})^{2}-(\sqrt{b})^{2}\\\\ &(\sqrt{b}-\sqrt{a})(\sqrt{b}+\sqrt{a})=(\sqrt{c}-\sqrt{b})(\sqrt{c}+\sqrt{b})\\\\ &\dfrac{\sqrt{b}-\sqrt{a}}{\sqrt{b}+\sqrt{c}}=\dfrac{\sqrt{c}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\\\\ &\dfrac{(\sqrt{b}+\sqrt{c})-(\sqrt{a}+\sqrt{c})}{\sqrt{b}+\sqrt{c}}=\dfrac{(\sqrt{a}+\sqrt{c})-(\sqrt{a}+\sqrt{b})}{\sqrt{a}+\sqrt{b}}\\\\ &1-\dfrac{\sqrt{a}+\sqrt{c}}{\sqrt{b}+\sqrt{c}}=\dfrac{\sqrt{a}+\sqrt{c}}{\sqrt{a}+\sqrt{b}}-1\\\\ &\text { Dividing both sides with } \sqrt{a}+\sqrt{c}\\\\ &\dfrac{1}{\sqrt{a}+\sqrt{c}}-\dfrac{1}{\sqrt{b}+\sqrt{c}}=\dfrac{1}{\sqrt{a}+\sqrt{b}}-\dfrac{1}{\sqrt{a}+\sqrt{c}}\\\\ \therefore\ &\dfrac{1}{\sqrt{b}+\sqrt{c}}, \dfrac{1}{\sqrt{a}+\sqrt{c}}, \dfrac{1}{\sqrt{a}+\sqrt{b}} \text { are in A.P } \end{aligned}$



13. If $a, b, c$ are in A.P., then prove that $a\left(\dfrac{1}{b}+\dfrac{1}{c}\right)$, $b\left(\dfrac{1}{c}+\dfrac{1}{a}\right)$, $c\left(\dfrac{1}{a}+\dfrac{1}{b}\right)$ are also in A.P.





$\begin{aligned} &a,\ b,\ c\ \text { ane in A.P.}\\\\ \therefore\ & b-a=c-b \\\\ \therefore\ & a+c=2 b \\\\ & a\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+c\left(\dfrac{1}{a}+\dfrac{1}{b}\right) \\\\ = & \dfrac{a}{b}+\dfrac{a}{c}+\dfrac{c}{a}+\dfrac{c}{b} \\\\ = & \dfrac{a+c}{b}+\dfrac{a}{c}+\dfrac{c}{a} \\\\ = & \dfrac{2 b}{b}+\dfrac{a^{2}+c^{2}}{a c}\\\\ = & 2+\dfrac{a^{2}+c^{2}}{a c} \\\\ = & 2+\dfrac{(a+c)^{2}-2 a c}{a c} \\\\ = & 2+\dfrac{(a+c)^{2}}{a c}-2 \\\\ = & \dfrac{(a+c)^{2}}{a c} \\\\ = & (a+c)\left(\dfrac{a+c}{a c}\right) \\\\ = & 2 b\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\\\\ \therefore\ & 2 b\left(\dfrac{1}{c}+\dfrac{1}{a}\right)=a\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+c\left(\dfrac{1}{a}+\dfrac{1}{b}\right) \\\\ & b\left(\dfrac{1}{c}+\dfrac{1}{a}\right)+b\left(\dfrac{1}{c}+\dfrac{1}{a}\right)=a\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+c\left(\dfrac{1}{a}+\dfrac{1}{b}\right) \\\\ & b\left(\dfrac{1}{c}+\dfrac{1}{a}\right)-a\left(\dfrac{1}{b}+\dfrac{1}{c}\right)=c\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-b\left(\dfrac{1}{c}+\dfrac{1}{a}\right) \\\\ \therefore\ & a\left(\dfrac{1}{b}+\dfrac{1}{c}\right), b\left(\dfrac{1}{c}+\dfrac{1}{a}\right), c\left(\dfrac{1}{a}+\dfrac{1}{b}\right) \text { are in A.P.} \end{aligned}$



14. If $a^{2}, b^{2}, c^{2}$ are in A.P., then prove that $\dfrac{1}{b+c}$, $\dfrac{1}{c+a}$, $\dfrac{1}{a+b}$ are also in A.P.





$\begin{aligned} &a^{2}, b^{2}, c^{2} \text { are in A.P.} \\\\ \therefore\ &b^{2}-a^{2}=c^{2}-b^{2} \\\\ &(b+a)(b-a)=(c+b)(c-b) \\\\ &\dfrac{b-a}{b+c}=\dfrac{c-b}{a+b} \\\\ &\dfrac{(b+c)-(c+a)}{b+c}=\dfrac{(c+a)-(a+b)}{a+b} \\\\ &1-\dfrac{c+a}{b+c}=\dfrac{c+a}{a+b}-1\\\\ &\text { Dividing both sides with } c+a \\\\ &\dfrac{1}{c+a}-\dfrac{1}{b+c}=\dfrac{1}{a+b}-\dfrac{1}{c+a} \\\\ \therefore\ &\dfrac{1}{b+c}, \dfrac{1}{c+a}, \dfrac{1}{a+b} \text { are in A.P.} \end{aligned}$



15. If $a^{2}$, $b^{2}$, $c^{2}$ are in A.P., then prove that $\dfrac{a}{b+c}$, $\dfrac{b}{c+a}$, $\dfrac{c}{a+b}$ are also in A.P.





$\begin{aligned} &a^{2}, b^{2}, c^{2} \text { are in A.P.} \\\\ \therefore\ &b^{2}-a^{2}=c^{2}-b^{2} \\\\ &(b+a)(b-a)=(c+b)(c-b) \\\\ &\dfrac{b-a}{b+c}=\dfrac{c-b}{a+b} \\\\ &\dfrac{(b+c)-(c+a)}{b+c}=\dfrac{(c+a)-(a+b)}{a+b} \\\\ &1-\dfrac{c+a}{b+c}=\dfrac{c+a}{a+b}-1\\\\ &\text { Dividing both sides with } c+a \\\\ &\dfrac{1}{c+a}-\dfrac{1}{b+c}=\dfrac{1}{a+b}-\dfrac{1}{c+a} \\\\ &\text { Multiplying both sides with } a+b+c,\\\\ &\dfrac{a+b+c}{c+a}-\dfrac{a+b+c}{b+c}=\dfrac{a+b+c}{a+b}-\dfrac{a+b+c}{c+a}\\\\ &\dfrac{c+a}{c+a}+\dfrac{b}{c+a}-\dfrac{a}{b+c}-\dfrac{b+c}{b+c}=\dfrac{a+b}{a+b}+\dfrac{c}{a+b}-\dfrac{c+a}{c+a}-\dfrac{b}{c+a}\\\\ &1+\dfrac{b}{c+a}-\dfrac{a}{b+c}-1=1+\dfrac{c}{a+b}-1-\dfrac{b}{c+a}\\\\ &\dfrac{b}{c+a}-\dfrac{a}{b+c}=\dfrac{c}{a+b}-\dfrac{b}{c+a}\\\\ &\dfrac{a}{b+c}, \dfrac{b}{c+a}, \dfrac{c}{a+b} \text { are in A.P. } \end{aligned}$



16. If the $m^{\text {th }}$ term of an A.P. is $\dfrac{1}{n}$ and $n^{\text {th }}$ term is $\dfrac{1}{m}$, then show that $u_{m n}=1$.





Let the first term and the common difference of the given A.P. be $a$ and $d$ respectively. $u_{m=} \dfrac{1}{n} $ $a+(m-1) d=\dfrac{1}{n}---(1) $ $u_{n}=\dfrac{1}{m} $ $a+(n-1) d=\dfrac{1}{m}---(2) $ $(1)-(2) \Rightarrow(m-n) d=\dfrac{1}{n}-\dfrac{1}{m}$ $(m-n) d=\dfrac{m-n}{m n}$ $d=\dfrac{1}{m n}$ $a+(m-1) \dfrac{1}{m n}=\dfrac{1}{n}$ $a=\dfrac{1}{n}-\dfrac{m-1}{m n}$ $\quad=\dfrac{m-m+1}{m n}$ $\quad=\dfrac{1}{m n}$ $u_{m n} =a+(m n-1) d $ $\quad\quad=\dfrac{1}{m n}+(m n-1) \dfrac{1}{m n} $ $\quad\quad=\dfrac{1}{m n}+1-\dfrac{1}{m n} $ $\quad\quad=1$



17. If the $p^{\text {th }}$ term of an A.P. is $q$ and the $q^{\text {th }}$ term is $p$, find its $n^{\text {th }}$ term in terms of $p, q$ and $n$.





$\begin{aligned} &\text{Let the first term}=a\\\\ &\text{the common difference}=d\\\\ &u_{p}=q \\\\ &a+(p-1) d=q---(1) \\\\ &u_{q}=p \\\\ &a+(q-1) d=p---(2) \\\\ &(1)-(2) \\\\ &(p-q) d=q-p \\\\ &(p-q) d=-(p-q)\\\\ &\therefore d=-1 \\\\ &\therefore a+(p-1)(-1)=q \\\\ &\begin{aligned} u_{n} &=a+(n-1) d \\\\ &=p+q-1+(n-1)(-1) \\\\ &=p+q-n \end{aligned} \end{aligned}$



18. If $\log _{10} 2, \log _{10}\left(2^{x}-1\right)$ and $\log _{10}\left(2^{x}+3\right)$ are three consecutive terms of an A.P., find the value of $x$.





$\begin{aligned} &\log _{10} 2, \log _{10}\left(2^{x}-1\right) \text { and } \log _{10}\left(2^{x}+3\right) \text { are in A.P.} \\\\ \therefore\ &\log _{10}\left(2^{x}-1\right)-\log _{10} 2=\log _{10}\left(2^{x}+3\right)-\log _{10}\left(2^{x}-1\right) \\\\ &\log _{10}\left(\frac{2^{x}-1}{2}\right)=\log _{10}\left(\frac{2^{x}+3}{2^{x}-1}\right) \\\\ &\frac{2^{x}-1}{2}=\frac{2^{x}+3}{2^{x}-1} \\\\ &\left(2^{x}-1\right)^{2}=2 \cdot 2^{x}+6 \\\\ &\left(2^{x}\right)^{2}-22^{x}+1=2 \cdot 2^{x}+6\\\\ &\left(2^{x}\right)^{2}-4 \cdot 2^{x}-5=0 \\\\ &\left(2^{x}+1\right)\left(2^{x}-5\right)=0 \\\\ & \text{For}\ 2^{x}=-1, \text { which is not possible for every } x \in \mathbb{R}.\\\\ &\text{For}\ 2^{x}=5, \\\\ \therefore\ & x=\log _{2} 5 \end{aligned}$