Arithmetic Progression : Problems and Solutions - Part (2)

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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 $x$.





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

Arithmetic Progression : Problems and Solutions - Part (1)

Arithmetic Progression

An arithmetic progression is a sequence in which the difference between two consecutive terms is a constant. ကိန်းစဉ်တစ်ခု၏ နီးစပ် (ကပ်လျက်) ကိန်းနှစ်လုံး ခြားနားခြင်းသည် ကိန်းသေဖြစ်လျှင် ၎င်းကိန်းစဉ်ကို arithmetic progression ဟုခေါ်သည်။ That constant is called the common difference of the progression. If $u_{1}, u_{2}, u_{3}, \ldots u_{n-1}, u_{n}$ is an A.P., then $u_{2}-u_{1}=u_{3}-u_{2}=\ldots=u_{n}-u_{n-1}=$ constant $u_{n}-u_{n-1}=d$ and $u_{n}=u_{n-1}+d$ where $\boldsymbol{d}$ is called the common difference. The $n^{\text {th }}$ term of an $A . P$ is given by $\begin{array}{|l|}\hline u_{n}=a+(n-1)d \\ \hline\end{array}$ where $u_{n}=n^{\text {th }}$ term, $a=$ first term (or) $u_{1}$, $d=$ common difference $n=$ number of terms

Exercises

1. In each of the following A.P., find
   (a) the common difference (b) the $10^{\text {th }}$ term (c) the $n^{\text {th }}$ term.
   (i) $1,3,5,7, \ldots$
   (ii) $10,9,8,7, \ldots$
   (iii) $1,2 \dfrac{1}{2}, 4,5 \dfrac{1}{2}, \ldots$
   (iv) $20,18,16,14, \ldots$
   (v)$-25,-20,-15,-10, \ldots$
   (vi) $-\dfrac{1}{8}$,$-\dfrac{1}{4}$,$-\dfrac{3}{8}$,$-\dfrac{1}{2}$, $\ldots$





(i) $1,3,5,7, \ldots$ is an A.P. $\quad$ (a) $d=3-1=2$ $\quad$ (b) $a=1, d=2$ $\hspace{1.2cm} u_{10} =a+9 d$ $\hspace{2cm} =1+9(2)$ $\hspace{2cm}=19$ $\quad$ (c) $ u_{n}\hspace{.3cm} =a+(n-1) d$ $\hspace{2cm}=1+(n-1) 2$ $\hspace{2cm} =1+2 n-2$ $\hspace{2cm}=2 n-1$ (ii) $10,9,8,7, \ldots$ is an A.P. $\quad$ (a) $d=9-10=-1$ $\quad$ (b) $a=10, d=-1$ $\hspace{1.2cm} u_{10} =a+9 d$ $\hspace{2cm} =10+9(-1)$ $\hspace{2cm}=1$ $\quad$ (c) $ u_{n}\hspace{.3cm} =a+(n-1) d$ $\hspace{2cm}=10+(n-1) (-1)$ $\hspace{2cm} =10-n+1$ $\hspace{2cm}=11-n$ (iii) $1,2\dfrac{1}{2},4, 5\dfrac{1}{2}, \ldots$ is an A.P. $\quad$ (a) $d=2\dfrac{1}{2}-1=1\dfrac{1}{2}=\dfrac{3}{2}$ $\quad$ (b) $a=1, d=\dfrac{3}{2}$ $\hspace{1.2cm} u_{10} =a+9 d$ $\hspace{2cm} =1+9\left(\dfrac{3}{2}\right)$ $\hspace{2cm}=\dfrac{29}{2}$ $\quad$ (c) $ u_{n}\hspace{.3cm} =a+(n-1) d$ $\hspace{2cm}=1+(n-1)\left(\dfrac{3}{2}\right)$ $\hspace{2cm} =1+\dfrac{3n}{2}-\dfrac{3}{2}$ $\hspace{2cm}=\dfrac{1}{2}(3n-1)$ (iv) $20,18,18,14, \ldots$ is an A.P. $\quad$ (a) $d=18-20=-2$ $\quad$ (b) $a=20, d=-2$ $\hspace{1.2cm} u_{10} =a+9 d$ $\hspace{2cm} =20+9(-2)$ $\hspace{2cm}=2$ $\quad$ (c) $ u_{n}\hspace{.3cm} =a+(n-1) d$ $\hspace{2cm}=20+(n-1) (-2)$ $\hspace{2cm} =20-2 n+2$ $\hspace{2cm}=22-2n$ (v) $-25,-20,-15,-10, \ldots$ is an A.P. $\quad$ (a) $d=-20-(-25)=5$ $\quad$ (b) $a=-25, d=5$ $\hspace{1.2cm} u_{10} =a+9 d$ $\hspace{2cm} =-25+9(5)$ $\hspace{2cm}=20$ $\quad$ (c) $ u_{n}\hspace{.3cm} =a+(n-1) d$ $\hspace{2cm}=-25+(n-1)5$ $\hspace{2cm} =-25+5n -5$ $\hspace{2cm}=5n-30$ (vi) $-\dfrac{1}{8},-\dfrac{1}{4},-\dfrac{3}{8},-\dfrac{1}{2}, \ldots$ is an A.P. $\quad$ (a) $d=-\dfrac{1}{4}-\left(-\dfrac{1}{8}\right)=- \dfrac{1}{8}$ $\quad$ (b) $a=-\dfrac{1}{8}, d=-\dfrac{1}{8}$ $\hspace{1.2cm} u_{10} =a+9 d$ $\hspace{2cm} =-\dfrac{1}{8}+9\left(-\dfrac{1}{8}\right)$ $\hspace{2cm}=-\dfrac{5}{4}$ $\quad$ (c) $ u_{n}\hspace{.3cm} =a+(n-1) d$ $\hspace{2cm}=-\dfrac{1}{8}+(n-1)\left(-\dfrac{1}{8}\right)$ $\hspace{2cm} =-\dfrac{1}{8}-\dfrac{n}{8}+\dfrac{1}{8}$ $\hspace{2cm}=\dfrac{n}{8}$



2. The $5^{\text {th }}$ term of an arithmetic progression is 10 while the $15^{\text {th }}$ term is 40 . Write down the first 5 terms of the A.P.





$u_{5}=10$ $a+4 d=10 \ldots (1)$ $u_{15}=40$ $a+14 d=40 \ldots (2)$ $(2)-(1)\Rightarrow 10 d=30$ $\therefore \ d=3$ Substituting $d=3$ in equation (1),we get $\quad\ a=-2$ $\therefore\ u_{1}=-2$ $\quad\ u_{2}=u_{1}+d=-2+3=1$ $\quad\ u_{3}=u_{2}+d=1+3=4$ $\quad\ u_{4}=u_{3}+d=4+3=7$ $\quad\ u_{5}=u_{4}+d=7+3=10$ $\therefore$ The first 5 terms are $-2,1,4,7,10$.



3. The $5^{\text {th }}$ and $10^{\text {th }}$ terms of an A.P. are 8 and $-7$ respectively. Find the $100^{\mathrm{th}}$ and $500^{\text {th }}$ terms of the A.P.





In an $A P$, $u_{5}=8$ $a+4 d=8 \ldots(1)$ $u_{10}=-7$ $a+9 d=-7 \ldots(2)$ $(2)-(1) \Rightarrow 5 d=-15$ $\hspace{2.6cm}d=-3$ $\therefore\ a+ 4(-3)=8 $ $\quad\ a= 20 $ $u_{100} =a+99 d $ $\quad\quad =20+99(-3) $ $\quad\quad =-277 $ $u_{500} =a+499 d $ $\quad\quad =20+499(-3) $ $\quad\quad =-1477$



4. The sixth term of an A.P. is 32 while the tenth term is 48 . Find the common difference and the $21^{\text {st }}$ term.





In an $A P$, $u_{6}=32$ $a+5 d=32 \ldots(1)$ $u_{10}=48$ $a+9 d=48 \ldots(2)$ $(2)-(1) \Rightarrow 4 d=16$ $\hspace{2.6cm}d=4$ $\therefore\ a+ 5(4)=32 $ $\quad\ a= 12 $ $u_{21} =a+20 d $ $\quad\ =12+20(4) $ $\quad\ =92 $



5. Which term of the A.P. $6,13,20,27, \ldots$ is $111 ?$





$6,13,20,27, \ldots,$ is an $A P .$ $\therefore\ a=6$ $d=13-6=7$ Let $u_{n}=111$ $a+(n-1) d=111$ $6+(n-1) 7=111$ $(n-1) 7=105$ $n-1=15$ $n=16$ $\therefore u_{16}=117$



6. If $u_{1}=6$ and $u_{30}=-52$ in an A.P., find the common difference.





$u_{1} =6$ $\therefore\ a =6$ $u_{30} =-52$ $a+29 d =-52$ $6+29 d=-52$ $29 d =-58$ $d =-2$



7. In an A.P., $u_{1}=3$ and $u_{7}=39$. Find (a) the first five terms of the A.P. (b) the $20^{\text {th }}$ term of the A.P.





$u_{1}=3$ $\therefore\ a=3$ $u_{7}=39$ $a+6 d=39$ $3+6 d=39$ $6 d=36$ $\therefore d=6$ $u_{1} =3 $ $u_{2} =u_{1}+d=3+6=9 $ $u_{3} =u_{2}+d=9+6=15 $ $u_{4} =u_{3}+d=15+6=21 $ $u_{5} =u_{4}+d=21+6=27 $ $u_{20} =a+19 d $ $\quad\ =3+19 \times 6 $ $\quad\ =117$



8. The four angles of a quadrilateral are in A.P. Given that the value of the largest is three times the value of the smallest angle, find the values of all four angles.





Let the four angles be $\alpha, \beta, \gamma$, and $\delta$. By the problem, $\alpha, \beta, r, \delta$ is an $A P .$ $\therefore\ \beta=\alpha+d$ $r=\alpha+2 d$ $\delta=\alpha+3 d$ where $d$ is a common difference. Since $\alpha+\beta+\gamma+\delta=360^{\circ}$ $4 \alpha+6 d=360^{\circ}$ $2 \alpha+3 d=180^{\circ}\ldots(1)$ By the prolblem, $\delta=3 \alpha$ $\alpha+3 d=3 \alpha$ $\therefore\ 2 \alpha-3 d=0 \ldots(2)$ $(1)+(2) \Rightarrow 4 \alpha =90^{\circ} $ $\alpha =45^{\circ} $ $(1)-(2) \Rightarrow 6 d =90^{\circ} $ $d =30^{\circ} $ $\therefore\ \alpha=45^{\circ} $ $\beta=45^{\circ}+30^{\circ} =75^{\circ} $ $\gamma =45^{\circ}+60^{\circ}=105^{\circ} $ $\delta=45^{\circ}+90^{\circ} =135^{\circ}$



9. If the $n^{\text {th }}$ term of an A.P. $2,3 \dfrac{7}{8}, 5 \dfrac{3}{4}, \ldots$ is equal to the $n^{\text {th }}$ term of an A.P. $187$ , $184 \dfrac{1}{4}$, $181 \dfrac{1}{2}$, $\ldots$, find $n$.





$2,3 \dfrac{7}{8}, 5 \dfrac{3}{4}, \ldots$ is an A.P. $a=2, d=3 \dfrac{7}{8}-2=1 \dfrac{7}{8}=\dfrac{15}{8}$ $u_{n}=a+(n-1) d$ $u_{n}=2+(n-1) \dfrac{15}{8}$ $187,184 \dfrac{1}{4}, 18-\dfrac{1}{2} \ldots$ is an A.P $a=187$ $d=184 \dfrac{1}{4}-187=-2 \dfrac{3}{4}=-\dfrac{11}{4}$ $u_{n}=a+(n-1) d$ $\quad\ =187+(n-1)\left(-\dfrac{11}{4}\right)$ $\quad\ =187-(n-1) \dfrac{n}{4}$ By the problem, $2+(n-1) \dfrac{15}{8}=187-(n-1) \dfrac{n}{4}$ $(n-1) \dfrac{15}{8}+(n-1) \dfrac{11}{4}=185$ $(n-1)\left(\dfrac{15}{8}+\dfrac{11}{4}\right)=185$ $(n-1) \dfrac{37}{8}=185$ $n-1=40$ $n=41$



10. Show that $\dfrac{1}{1+x}, \dfrac{1}{1-x^{2}}, \dfrac{1}{1-x}$ are three consecutive terms of an A.P.





$\begin{aligned} \frac{1}{1-x^{2}}-\frac{1}{1+x} &=\frac{1}{1-x^{2}}-\frac{1}{1+x} \times \frac{1-x}{1-x^{2}} \\\\ &=\frac{1}{1-x^{2}}-\frac{1-x}{1-x^{2}} \\\\ &=\frac{x}{1-x^{2}} \\\\ \frac{1}{1-x}-\frac{1}{1-x^{2}}&=\frac{1}{1-x} \times \frac{1+x}{1+x}-\frac{1}{1-x^{2}} \\\\ &=\frac{1+x}{1-x^{2}}-\frac{1}{1-x^{2}} \\\\ &=\frac{x}{1-x^{2}}\\\\ \therefore\ \frac{1}{1-x^{2}}-\frac{1}{1+x}&=\frac{1}{1-x}-\frac{1}{1-x^{2}} \\\\ \therefore\ \frac{1}{1+x},\ \frac{1}{1-x^{2}},\ & \frac{1}{1-x} \text { is an A.P.} \end{aligned}$



11. The first three terms of an A.P. are $4 p^{2}-10,8 p$ and $4 p+3$ respectively. Find the two possible values of $p .$ If $p$ is positive and that the $n^{\text {th }}$ term of the progression is $-93$, find the value of $n$.





$4 p^{2}-10,8 p, 4 p+3$ are the first three terms of an A.P. $\therefore 8 p-\left(4 p^{2}-10\right)=4 P+3-8 p$ $4 p^{2}-12 p-7=0$ $(2 p+1)(2 p-7)=0$ $\quad p=-\dfrac{1}{2}$ or $p=\dfrac{7}{2}$ If $p>0, \quad p=\dfrac{7}{2}$ $\therefore a=4\left(\dfrac{7}{2}\right)^{2}-10=39$ $u_{2}=8\left(\dfrac{7}{2}\right)=28$ $d=28-39=-11$ $u_{n}=-93$ $a+(n-1) d=-93$ $39+(n-1)(-11)=-93$ $(n-1)(-11)=-132$ $n-1=12$ $n=13$



12. The $5^{\text {th }}$ term and $8^{\text {th }}$ terms of an A.P. are $x$ and $y$ respectively. Show that the $20^{\text {th }}$ term is $5 y-4 x$.





$\quad u_{5}=x$ $\quad a+4 d=x$ $\quad u_{8}=y$ $\quad a+7 d=y$ $\quad 5 y-4 x$ $=5 a+35 d-4 a-16 d$ $=a+19 d$ $=u_{20}$ $\therefore u_{20}=5 y-4 x$



13. Given that $\dfrac{1}{b+c}, \dfrac{1}{c+a}, \dfrac{1}{a+b}$ are three consecutive terms of an A.P., show also that $a^{2}, b^{2}$ and $c^{2}$ are three consecutive terms of an A.P.





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



14. A certain A.P. has 25 terms. The last three terms are $\dfrac{1}{x-4}, \dfrac{1}{x-1}$ and $\dfrac{1}{x}$, Calculate the value $x$, and the middle term of that progression.





$ \dfrac{1}{x-1}-\dfrac{1}{x-4}=\dfrac{1}{x}-\dfrac{1}{x-1} $ $ \dfrac{x-4-x+1}{(x-1)(x-4)}=\dfrac{x-1-x}{x(x-1)} $ $ \dfrac{-3}{x-4}=\dfrac{-1}{x} $ $ \therefore\ 3 x=x-4 $ $ \quad\ 2 x=-4 $ $ \quad\ x=-2 $ $\therefore\ u_{23}=\dfrac{1}{x-4}=-\dfrac{1}{6} $ $ \quad\ u_{24}=\dfrac{1}{x-1}=-\dfrac{1}{3} $ $ \quad\ u_{25}=\dfrac{1}{x}=-\dfrac{1}{2}$ $\quad\ d=-\dfrac{1}{3}+\dfrac{1}{6}=-\dfrac{1}{6}$ $\quad\ u_{23}=-\dfrac{1}{6}$ $\quad\ a+22 d=-\dfrac{1}{6}$ $\therefore\ a=\dfrac{21}{6} $ $\therefore\ \text { middle term } =u_{13} $ $\hspace{3.2cm}=a+12 d $ $\hspace{3.2cm}=\dfrac{21}{6}+12\left(-\dfrac{1}{6}\right) $ $\hspace{3.2cm}=\dfrac{3}{2}$ $\hspace{3.5cm}\mathrm{OR}$ $\quad\ \text { middle term } =\dfrac{u_{1}+u_{25}}{2}$ $\hspace{3.2cm}=\dfrac{\dfrac{21}{6}-\dfrac{1}{2}}{2}$ $\hspace{3.2cm}=\dfrac{3}{2} $



15. Given that $x^{2},(8 x+1)$ and $(7 x+2)$, where $x \neq 0$, are the $2^{\text {nd }}, 4^{\text {th }}$ and $6^{\text {th }}$ terms respectively of an A.P. Find the value of $x$, the common difference and the first term.





$x^{2}, 8 x+1,7 x+2$ are $2^{\text{nd}}, 4^{\text{th}}$ and $6^{\text{th}}$ terms of an A.P. $\therefore\ 8 x+1-x^{2}=7 x+2-(8 x+1)$ $\quad\ x^{2}-9 x=0$ $\quad\ x(x-9)=0$ Since $x \neq 0, x-9=0,$ $\therefore\ x=9 .$ $\therefore\ u_{2}=81$ $\quad\ a+d=81$ ---(1) $\quad\ u_{4}=73$ $\quad\ a+3 d=73$ --- (2) $(2)-(1) \Rightarrow 2 d=-8$ $\hspace{2.6cm} d=-4$ $\therefore\ a-4=81 $ $\quad\ a=85$



16. If $\dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}$ are in A.P., express $b$ in terms of $a$ and $c$.





$\dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}$ ane in A.P. $\dfrac{1}{b}-\dfrac{1}{a}=\dfrac{1}{c}-\dfrac{1}{b}$ $\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}$ $\dfrac{2}{b}=\dfrac{a+c}{a c}$ $b=\dfrac{2 a c}{a+c}$



17. If $m$ times the $m^{\text {th }}$ term of an A.P. is equal to $n$ times the $n^{\text {th }}$ term where $m \neq n$ find $(m+n)^{\text {th }}$ term of the progression.





Let $a$ and $d$ be the first term and the common difference of given A.P. By the problem, $\quad\ m u_{m}=n u_{n}$ $\quad\ m(a+(m-1) d)=n(a+(n-1) d)$ $\quad\ m a+\left(m^{2}-n\right) d=n a+\left(n^{2}-n\right) d$ $\quad\ m a-n a+\left(m^{2}-n^{2}-m+n\right) d=0$ $\quad\ (m-n) a+[(m+n)(m-n)-(m-n)] d=0$ $\quad\ (m-n)[a+(m+n-1) d]=0$ $\quad\ $ Since $m \neq n, m-n \neq 0$ $\therefore\ a+(m+n-1) d=0$ $\therefore\ u_{m+n}=0$



18. If $a^{2}+2 b c, b^{2}+2 a c, c^{2}+2 a b$ are in A.P., show that $\dfrac{1}{b-c}$, $\dfrac{1}{c-a}$, $\dfrac{1}{a-b}$ are in A.P.





$a^{2}+2 b c, b^{2}+2 a c, c^{2}+2 a b$ are in A.P. $\therefore\ b^{2}+2 a c-a^{2}-2 b c=c^{2}+2 a b-b^{2}-2 a c$ $\quad\ b^{2}-a^{2}+a c(a-b)=c^{2}-b^{2}+2 a(b-c)$ $\quad\ (b-a)(b+a)-2 c(b-a)=(c-b)(c+b)-2 a(c-b)$ $\quad\ (b-a)(b+a-2 c)=(c-b)(c+b-2 a)$ $\quad\ -(a-b)(a+b-2 c)=-(b-c)(b+c-2 a)$ $\quad\ (a-b)(2 c-a-b)=(b-c)(2 a-b-c)$ $\quad\ (a-b)[(c-a)+(c-b)]=(b-c)[(a-b)+(a-c)]$ $\quad\ (a-b)[(c-a)-(b-c)]=(b-c)[(a-b)-(c-a)]$ $\quad\ (a-b)(c-a)-(a-b)(b-c)=(b-c)(a-b)-(b-c)(c-a)$ Dividing both sides with $(a-b)(b-c)(c-a)$, $\quad\ \dfrac{1}{b-c}-\dfrac{1}{c-a}=\dfrac{1}{c-a}-\dfrac{1}{a-b}$ $\quad\ \dfrac{1}{c-a}-\dfrac{1}{b-c}=\dfrac{1}{a-b}-\dfrac{1}{c-a}$ $\therefore\ \dfrac{1}{b-c}, \dfrac{1}{c-a}, \dfrac{1}{a-b}$ are in A.P.

Grade 10: Exercise (4.6) - Solution

A real valued function is one-to-one if every horizontal line intersects the graph of the function at most one point.

1. Determine whether each of the following function is a one-to-one function or not. If it is not one-to-one, explain why not.

(a) It is a one to one function.

(b) It is not a one to one function because some horizontal lines intersect the graph of the function more than one point.

(c) It is a one to one function.

(d) It is not a one to one function because some horizontal lines intersect the graph of the function more than one point.

(c) It is a one to one function.

(c) It is a one to one function.

2. Draw the graph of the each given function and determine whether each is a one-to-one function or not.
   
   (a) $f(x)=3 x+2$
   
   $\begin{array}{|c||c|c|c|c|c|c|c|} \hline x & \ldots & -2 & -1 & 0 & 1 & 2 & \ldots \\ \hline f(x) & \cdots & -4 & -1 & 2 & 5 & 8 & \ldots \\ \hline \end{array}$
   
   

   It is a one to one function.

   (b) $f(x)=x-3$
   
   $\begin{array}{|c||c|c|c|c|c|c|c|} \hline x & \ldots & -2 & -1 & 0 & 1 & 2 & . \\ \hline f(x) & \cdots & -5 & -4 & -3 & 2 & 1 & \cdots \\ \hline \end{array}$
   
   

   It is a one to one function.

   (c) $f(x)=4 x^{2}$
   
   $\begin{array}{|c||c|c|c|c|c|c|c|} \hline x & \cdots & -2 & -1 & 0 & 1 & 2 & \cdots \\ \hline f(x) & \cdots & 16 & 4 & 0 & 4 & 16 & \cdots \\ \hline \end{array}$
   
   

   It is not a one to one function.

   (d) $f(x)=2|x|$
   
   $\begin{array}{|c||c|c|c|c|c|c|c|} \hline x & \cdots- & -2 & -1 & 0 & 1 & 2 & \cdots \\ \hline f(x) & \cdots & 4 & 2 & 0 & 2 & 4 & \cdots \\ \hline \end{array}$
   
   

   It is not a one to one function.

   (e) $f(x)=\dfrac{2 x+3}{x+2}$
   
   $\begin{aligned} f(x) &=\dfrac{2 x+3}{x+2} \\\\ &=\dfrac{2 x+4-1}{(x+2)} \\\\ &=\dfrac{-1+2(x+2)}{(x+2)} \\\\ &=\dfrac{-1}{x+2}+2\\\\ &\text { horizontal arymptote: } y=2 \\\\ &\text { vertical asymptote : } x=-2 \\\\ &x=0, y=\frac{3}{2} \\\\ &y \text { -intercept }=\left(0, \frac{3}{2}\right) \\\\ &y=0, x=-\frac{3}{2} \\\\ &x \text { -intercept }=\left(-\frac{3}{2}, 0\right) \end{aligned}$
   
   

   It is a one to one function.

   (f) $f(x)=4 x^{2}\quad (0 \le x \le 4)$
   
   $\begin{array}{|c||c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 & 4 \\ \hline f(x) & 0 & 4 & 16 & 36 & 64 \\ \hline \end{array}$
   
   

   It is a one to one function.

   (g) $f(x)=\sqrt{x}\quad (x \ge 0)$
   
   $\begin{array}{|c||c|c|c|c|c|c|} \hline x & 0 & 1 & 4 & 9 & 25 & \ldots \\ \hline f(x) & 0 & 1 & 2 & 3 & 5 & \cdots \\ \hline \end{array}$
   
   

   It is a one to one function.