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 $x$.





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

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