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.

Sequences and Series

Sequence

A sequence is a function whose domain is either the set of all or part of the natural numbers. The values of the function (images) are called the terms of the of the sequence. 
Sequence ဆိုတာ Function တစ်ခုဖြစ်ပါတယ် ၎င်း၏ domain မှာ set of natural numbers (သဘာဝကိန်းများ ပါဝင်သော အစု) ဖြစ်ပါသည်။ Image များကိုတော့ term လို့ ခေါ်ပါတယ်။

ပုံတွင်

1 ၏ image သည် 1 ဖြစ်လို့ ⇒ f(1) = 1
2 ၏ image သည် 4 ဖြစ်လို့ ⇒ f(2) = 4
3 ၏ image သည် 9 ဖြစ်လို့ ⇒ f(3) = 9 
4 ၏ image သည် 16 ဖြစ်လို့ ⇒ f(4) = 16
⋮  

n ၏ image သည် n² ဖြစ်လို့ ⇒ f(n) = n² သတ်မှတ်နိုင်ပါတယ်။ အထက်ပါ သတ်မှတ်ချက်ဟာ function ကို လေ့လာခဲ့စဉ်က သတ်မှတ်သော functional notation ဖြစ်ပါတယ်။ Sequence မှာတော့ functional notation အစား အခြား notation ကို ပြောင်းသုံးပါမယ်။ ဒါ့ကြောင့်

$ \displaystyle f(1)$ အစား $ \displaystyle u_1 = 1^{\text{st}}\ \text{term}$

$ \displaystyle f(2)$ အစား $ \displaystyle u_2= 2^{\text{nd}}\ \text{term}$

$ \displaystyle f(3)$ အစား $ \displaystyle u_3= 3^{\text{rd}}\ \text{term}$

 ⋮
 $ \displaystyle f(n)$ အစား $ \displaystyle u_n= n^{\text{th}}\ \text{term}$ လို့ ပြောင်းသုံးပါမယ်။

The value of the function corresponding to the number n of the domain is called the n^th term or the general term of the sequence.

Domain ထဲရှိ အစုဝင် n နှင့် ဆက်စက်နေသော image ($ \displaystyle u_n$) ကို n^th term ဟု ခေါ်ပါတယ်။ ၎င်းကို sequence တစ်ခု ၏ general term လို့ လဲ သတ်မှတ်ပါတယ်။ Sequence ဖြစ်ပေါ်လာပုံ သေနည်း (Rule of Formation) လို့ ပြောနိုင်ပါတယ်။

Progression

When terms of a sequence are written under specific conditions, then the sequence is called a progression.

sequence တစ်ခုကို ဖြစ်ပေါ်လာပုံနည်း စနစ်တစ်ခု နဲ့ တည်ဆောက်ထားရင် ... တစ်နည်း general term တည်ဆောက်နိုင်နိုင်ရင် progression လို့ ခေါ်ပါတယ်။

progression တိုင်းက sequence များ ဖြစ်ကြပေမယ့် sequence တိုင်းကတော့ progression တွေ မဟုတ်ကြပါဘူး...

(1) သဘာဝကိန်းများ
(2) နှစ်ထပ်ကိန်းတိများ
(3) သုဒ္ဓကိန်းများ 

အထက်ပါသတ်မှတ်ချက်တွေဟာ sequence တွေ ဖြစ်တယ်လို့ ဆိုနိုင်ပါတယ်.. ဒါပေမယ့် သဘာဝကိန်းနဲ့ နှစ်ထပ်ကိန်းတိများ ဆိုတဲ့ သတ်မှတ်ချက်အတွက် general term (formula) ရှာလို့ရပေမယ့် သုဒ္ဓကိန်း တွေအတွက်တော့ general term ရှာလို့မရပါဘူး ဒါ့ကြောင့် သုဒ္ဓကိန်းများ ဖြစ်တဲ့..

2, 3, 5, 7, 11, ... ဆိုတာ sequnce တစ်ခုလို့ ဆိုနိုင်သော်လည်း progression တစ်ခုတော့ မဟုတ်ပါဘူး။ အလားတူးပါပဲ.. 1976 မှ 2019 အထိ ချဲဂဏန်း (ပေါက်ဂဏန်းများ) ဆိုတာ sequence တစ်ခု ဖြစ်ပါတယ်။ ဒါပေမယ့် progression တော့ မဟုတ်ပါဘူး။ ဘယ်နှစ်ရဲ့ ဘယ်အကြိမ်မှာ ဘာဂဏန်း ထွက်လာမယ်လို့ အတိအကျ မသတ်မှတ်နိုင်လို့ပါပဲ။ 

Geometric Progression : Problem and Solutions

1.        If $ \displaystyle a, b, c, d$ are in $ \displaystyle G.P.,$ prove that $ \displaystyle a+b, b+c, c+d$ are also in $ \displaystyle G.P.$

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$ \displaystyle a, b, c, d$ are in $ \displaystyle G.P.$ Let $ \displaystyle r$ be the common ratio. $ \displaystyle \begin{array}{l}\therefore \ a=a,\ b=ar,\ c=a{{r}^{2}},\ d=a{{r}^{3}}\\\\\therefore \ \displaystyle \frac{{b+c}}{{a+b}}=\displaystyle \frac{{ar+a{{r}^{2}}}}{{a+ar}}\\\\\therefore \ \displaystyle \frac{{b+c}}{{a+b}}=\displaystyle \frac{{r\left( {ar+ar} \right)}}{{a+ar}}\\\\\therefore \ \displaystyle \frac{{b+c}}{{a+b}}=r\\\\\text{Again},\\\\\ \ \ \displaystyle \frac{{c+d}}{{b+c}}=\displaystyle \frac{{a{{r}^{2}}+a{{r}^{3}}}}{{ar+a{{r}^{2}}}}\\\\\therefore \ \displaystyle \frac{{c+d}}{{b+c}}=\displaystyle \frac{{r\left( {ar+a{{r}^{2}}} \right)}}{{ar+a{{r}^{2}}}}\\\\\therefore \ \displaystyle \frac{{c+d}}{{b+c}}=r\\\\\therefore \displaystyle \frac{{b+c}}{{a+b}}=\displaystyle \frac{{c+d}}{{b+c}}\end{array}$ $ \displaystyle \therefore \ a+b, b+c, c+d$ are also in $ \displaystyle G.P.$

2.         Find three numbers in $ \displaystyle G.P.$ whose sum is $ \displaystyle 13$ and the sum of whose squares is $ \displaystyle 91$.

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Let the three numbers in $ \displaystyle G.P.$ be $ \displaystyle a, ar, ar^2$ By the problem, $ \displaystyle a+ar+ar^2=13$ $ \displaystyle a(1+r+r^2)=13---(1)$ $ \displaystyle a^2+(ar)^2+(ar^2)^2=91$ $ \displaystyle \therefore \ a^2+a^2r^2+a^2r^4=91$ $ \displaystyle \therefore \ a^2(1+r^2+r^4)=91---(2)$ Squaring both sides of equation $ \displaystyle (1),$ $ \displaystyle a^2(1+r+r^2)^2=169$ $ \displaystyle \therefore a^2(1+2r+3{{r}^{2}}+2{{r}^{3}}+{{r}^{4}})=169$ $ \displaystyle \begin{array}{l}\therefore {{a}^{2}}\left[ {1+2r+3{{r}^{2}}+2{{r}^{3}}+{{r}^{4}}} \right]=169\\\\\therefore {{a}^{2}}\left[ {1+{{r}^{2}}+{{r}^{4}}+2r+2{{r}^{2}}+2{{r}^{3}}} \right]=169\\\\\therefore {{a}^{2}}\left[ {\left( {1+{{r}^{2}}+{{r}^{4}}} \right)+2r\left( {1+r+{{r}^{2}}} \right)} \right]=169\\\\\therefore {{a}^{2}} {\left( {1+{{r}^{2}}+{{r}^{4}}} \right)+2r\cdot {{a}^{2}}\left( {1+r+{{r}^{2}}} \right)}=169\\\\\therefore 91+2ar\cdot 13=169\\\\\therefore 26ar=78\\\\\therefore ar=3\\\\\therefore a=\displaystyle \frac{3}{r}\end{array}$ Substituting $ \displaystyle a=\frac{3}{r}$ in equation $ \displaystyle (1),$ $ \displaystyle \begin{array}{l}\ \ \frac{3}{r}(1+r+{{r}^{2}})=13\\\\\therefore 3+3r+3{{r}^{2}}=13r\\\\\therefore 3{{r}^{2}}-10r+3=0\\\\\therefore (3r-1)(r-3)=0\end{array}$ $ \displaystyle \therefore r=\frac{1}{3}$ (or) $ \displaystyle r=3.$ When $ \displaystyle r=\frac{1}{3}, a= 9.$ Therefore the numbers are 9, 3 and 1. When $ \displaystyle r=3, a= 1.$ Therefore the numbers are 1, 3 and 9.

3.        The product of three consecutive terms of a $ \displaystyle G.P.$ is $ \displaystyle 8.$ The sum of product of these terms taken in pairs is $ \displaystyle 14.$ Find the numbers.

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Let the three consecutive terms in $ \displaystyle G.P.$ be $ \displaystyle \frac{a}{r}, a, ar$ By the problem, $ \displaystyle \frac{a}{r}\cdot a\cdot ar=8$ $ \displaystyle \begin{array}{l}\ \ \ \displaystyle \frac{a}{r}\cdot a\cdot ar=8\\\\\therefore \ {{a}^{3}}=8\\\\\therefore \ a=2\\\\\ \ \text{Again},\\\\\ \ \displaystyle \frac{a}{r}\cdot a+a\cdot ar+\displaystyle \frac{a}{r}\cdot a=14\\\\\therefore \displaystyle \frac{{{{a}^{2}}}}{r}+{{a}^{2}}r+{{a}^{2}}=14\\\\\therefore {{a}^{2}}\left( {\displaystyle \frac{1}{r}+r+1} \right)=14\\\\\therefore 4\left( {\displaystyle \frac{1}{r}+r+1} \right)=14\\\\\therefore \displaystyle \frac{1}{r}+r+1=\displaystyle \frac{7}{2}\\\\\therefore 2+2{{r}^{2}}+2r=7r\\\\\therefore 2{{r}^{2}}-5r+2=0\\\\\therefore (2r-1)(r-2)=0\\\\\therefore \ r=\displaystyle \frac{1}{2}\ (\text{or})\ r=2\end{array}$ Therefore the numbers are 4, 2 , 1 or 1, 2, 4.

4.         The sum of two positive numbers is $ \displaystyle 6$ times their geometric mean. Show that the numbers are in the ratio $ \displaystyle \left( {3+2\sqrt{2}} \right):\left( {3-2\sqrt{2}} \right)$.

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Let the two positive numbers be $ \displaystyle a$ and $ \displaystyle b.$ $ \displaystyle G.M.$ between $ \displaystyle a$ and $ \displaystyle b.$ = $ \displaystyle \sqrt{ab}$ By the problem, $ \displaystyle \begin{array}{l}\ \ \ a+b=6\sqrt{{ab}}\\\\\therefore \ {{a}^{2}}+2ab+{{b}^{2}}=36ab\\\\\therefore \ {{a}^{2}}+{{b}^{2}}=34ab\\\\\therefore \ \displaystyle \frac{{{{a}^{2}}+{{b}^{2}}}}{{ab}}=34\\\\\therefore \ \displaystyle \frac{a}{b}+\displaystyle \frac{b}{a}=34\\\\\text{Let}\ \displaystyle \frac{a}{b}=x.\\\\\therefore \displaystyle \frac{b}{a}=\displaystyle \frac{1}{x}\\\\\therefore x+\displaystyle \frac{1}{x}=34\\\\\therefore {{x}^{2}}-34x+1=0\\\\\therefore {{x}^{2}}-34x=-1\\\\\therefore {{x}^{2}}-34x+289=288\\\\\therefore {{(x-17)}^{2}}=288\\\\\therefore x-17=12\sqrt{2}\\\\\therefore x=17+12\sqrt{2}\\\\\therefore x=9+12\sqrt{2}+8\\\\\therefore x={{3}^{2}}+2\left( 3 \right)\left( {2\sqrt{2}} \right)+{{\left( {2\sqrt{2}} \right)}^{2}}\\\\\therefore x={{\left( {3+2\sqrt{2}} \right)}^{2}}\\\\\therefore x=\displaystyle \frac{{{{{\left( {3+2\sqrt{2}} \right)}}^{2}}}}{{9-8}}\\\\\therefore x=\displaystyle \frac{{{{{\left( {3+2\sqrt{2}} \right)}}^{2}}}}{{{{3}^{2}}-{{{\left( {2\sqrt{2}} \right)}}^{2}}}}\\\\\therefore x=\displaystyle \frac{{\left( {3+2\sqrt{2}} \right)\left( {3+2\sqrt{2}} \right)}}{{\left( {3+2\sqrt{2}} \right)\left( {3-2\sqrt{2}} \right)}}\\\\\therefore x=\displaystyle \frac{{3+2\sqrt{2}}}{{3-2\sqrt{2}}}\\\\\therefore \displaystyle \frac{a}{b}=\displaystyle \frac{{3+2\sqrt{2}}}{{3-2\sqrt{2}}}\end{array}$

5.         The sum of an infinite $ \displaystyle G.P$ is $ \displaystyle 57$ and the sum of their cubes is $ \displaystyle 9747,$ find the fourth term.

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Let the first term be $ \displaystyle a$ and the common ratio be $ \displaystyle r$. $ \displaystyle \therefore$ Given $ \displaystyle G.P$ is $ \displaystyle a, ar, ar^2,...$ Let sum to infinity be $ \displaystyle S$. $ \displaystyle \therefore S =57$ Since $ \displaystyle S=\frac{a}{{1-r}},$ $ \displaystyle \frac{a}{{1-r}}=57---(1)$ When each terms are cubed, given $ \displaystyle G.P$ becomes ... $ \displaystyle a^3, a^3r^3, a^3r^6,...$ It is also a $ \displaystyle G.P$ with the first term $ \displaystyle a^3$ and the common ratio $ \displaystyle r^3.$ Let the sum to infinity of that $ \displaystyle G.P$ be $ \displaystyle S^{*}$ $ \displaystyle \therefore S^{*} =9747$ Since $ \displaystyle {{S}^{*}}=\frac{{{{a}^{3}}}}{{1-{{r}^{3}}}},$ $ \displaystyle \frac{{{{a}^{3}}}}{{1-{{r}^{3}}}}=9747---(2)$ Cubing equation $ \displaystyle (1)$ on both sides, $ \displaystyle \ \ \ \frac{{{{a}^{3}}}}{{{{{\left( {1-r} \right)}}^{3}}}}={{57}^{3}}---(3)$ Dividing equation $ \displaystyle (3)$ by equation $ \displaystyle (2),$ $ \displaystyle \begin{array}{l}\ \ \ \displaystyle \frac{{\displaystyle \frac{{{{a}^{3}}}}{{{{{\left( {1-r} \right)}}^{3}}}}}}{{\displaystyle \frac{{{{a}^{3}}}}{{1-{{r}^{3}}}}}}=\displaystyle \frac{{{{{57}}^{3}}}}{{9747}}\\\\\therefore \ \ \displaystyle \frac{{{{a}^{3}}}}{{{{{\left( {1-r} \right)}}^{3}}}}\times \displaystyle \frac{{1-{{r}^{3}}}}{{{{a}^{3}}}}=19\\\\\therefore \ \ \displaystyle \frac{{\left( {1-r} \right)\left( {1+r+{{r}^{2}}} \right)}}{{\left( {1-r} \right)\left( {1-2r+{{r}^{2}}} \right)}}=19\\\\\therefore \ \ \displaystyle \frac{{1+r+{{r}^{2}}}}{{1-2r+{{r}^{2}}}}=19\\\\\therefore \ \ 1+r+{{r}^{2}}=19-38r+19{{r}^{2}}\\\\\therefore \ \ 18{{r}^{2}}-39r+18=0\\\\\therefore \ \ 6{{r}^{2}}-13r+6=0\\\\\therefore \ \ (3r-2)(2r-3)=0\\\\\therefore \ \ r=\displaystyle \frac{2}{3}\ (\text{or})\ r=\displaystyle \frac{3}{2}\end{array}$ Since the sum to infinity exists, |$ \displaystyle r$|<1. $ \displaystyle \begin{array}{l}\therefore \ \ r=\displaystyle \frac{2}{3}\\\\\therefore \ \ \displaystyle \frac{a}{{1-\displaystyle \frac{2}{3}}}=57\\\\\therefore \ \ a=19\\\\\therefore \ \ {{u}_{4}}=a{{r}^{3}}\\\\\therefore \ \ {{u}_{4}}=19\times \displaystyle \frac{8}{{27}}\\\\\therefore \ \ {{u}_{4}}=\displaystyle \frac{{152}}{{27}}\end{array}$

6.         The sum of an infinite $ \displaystyle G.P$ is $ \displaystyle x$ and the sum of their squares is $ \displaystyle y,$ find the common ratio of that $ \displaystyle G.P$ in terms of $ \displaystyle x$ and $ \displaystyle y$.

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Let the first term be $ \displaystyle a$ and the common ratio be $ \displaystyle r$. $ \displaystyle \therefore$ Given $ \displaystyle G.P$ is $ \displaystyle a, ar, ar^2,...$ By the problem, the sum to infinity of the progression is $ \displaystyle x$. $ \displaystyle \therefore \ \ \frac{a}{{1-r}}=x$ $ \displaystyle \therefore \ \ a=x(1-r)$ When each terms are squared, given $ \displaystyle G.P$ becomes, $ \displaystyle a^2, a^2r^2, a^2r^4,...$ It is also a $ \displaystyle G.P$ with the first term $ \displaystyle a^2$ and the common ratio $ \displaystyle r^2.$ By the problem, the sum to infinity of above $ \displaystyle G.P$ is $ \displaystyle y$. $ \displaystyle \begin{array}{l}\therefore \ \ \displaystyle \frac{{{{a}^{2}}}}{{1-{{r}^{2}}}}=y\\\\\therefore \ \ \displaystyle \frac{a}{{1-r}}\times \displaystyle \frac{a}{{1+r}}=y\\\\\therefore \ \ x\times \displaystyle \frac{a}{{1+r}}=y\\\\\therefore \ \ x\times \displaystyle \frac{{x(1-r)}}{{1+r}}=y\\\\\therefore \ \ \displaystyle \frac{{1+r}}{{1-r}}=\displaystyle \frac{{{{x}^{2}}}}{y}\\\\\therefore \ \ \displaystyle \frac{{\left( {1+r} \right)+\left( {1-r} \right)}}{{\left( {1+r} \right)-\left( {1-r} \right)}}=\displaystyle \frac{{{{x}^{2}}+y}}{{{{x}^{2}}-y}}\\\ \ \ \ \left[ {\because \ \text{componendo and dividendo}} \right]\\\\\therefore \ \ \displaystyle \frac{2}{{2r}}=\displaystyle \frac{{{{x}^{2}}+y}}{{{{x}^{2}}-y}}\\\\\therefore \ \ r=\displaystyle \frac{{{{x}^{2}}-y}}{{{{x}^{2}}+y}}\end{array}$

7.         Given that $ \displaystyle a, b, c$ are in $ \displaystyle A.P.$ and $ \displaystyle a^2, b^2, c^2$ are in $ \displaystyle G.P.$ If $ \displaystyle a<b<c$ and $ \displaystyle a+b+c=\frac{3}{2},$ Find the value of $ \displaystyle a, b$ and $ \displaystyle c.$

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$ \displaystyle a, b, c$ are in $ \displaystyle A.P.$ (given) Let the common difference be $ \displaystyle d$ where $ \displaystyle d\ne 0$. $ \displaystyle \therefore \ a=b-d\ \operatorname{and}\ c=b+d$ By the problem, $ \displaystyle \begin{array}{l}\ \ \ \ a+b+c=\displaystyle \frac{3}{2}\\\\\therefore \ \ b-d+b+b+d=\displaystyle \frac{3}{2}\\\\\therefore \ \ 3b=\displaystyle \frac{3}{2}\\\\\therefore \ \ b=\displaystyle \frac{1}{2}\\\\\therefore \ a=b-\displaystyle \frac{1}{2}\ \operatorname{and}\ c=b+\displaystyle \frac{1}{2}\end{array}$ $ \displaystyle a^2, b^2, c^2$ are in $ \displaystyle G.P.$ (given) $ \displaystyle \therefore \ {{\left( {\frac{1}{2}-d} \right)}^{2}},{\displaystyle \frac{1}{4}},{{\left( {\frac{1}{2}+d} \right)}^{2}}$ are in $ \displaystyle G.P.$ $ \displaystyle \begin{array}{l}\therefore \displaystyle \frac{{\displaystyle \frac{1}{4}}}{{{{{\left( {\displaystyle \frac{1}{2}-d} \right)}}^{2}}}}=\displaystyle \frac{{{{{\left( {\displaystyle \frac{1}{2}+d} \right)}}^{2}}}}{{\displaystyle \frac{1}{4}}}\\\\\therefore {{\left( {\displaystyle \frac{1}{4}-{{d}^{2}}} \right)}^{2}}=\displaystyle \frac{1}{{16}}\\\\\therefore \displaystyle \frac{1}{4}-{{d}^{2}}=\displaystyle \frac{1}{4}\ (\text{or})\displaystyle \frac{1}{4}-{{d}^{2}}=-\displaystyle \frac{1}{4}\\\\\therefore {{d}^{2}}=0\ (\text{or})\ {{d}^{2}}=\displaystyle \frac{1}{2}\end{array}$ But $ \displaystyle {{d}^{2}}=0$ is impossible, $ \displaystyle \begin{array}{l}\therefore \ {{d}^{2}}=\displaystyle \frac{1}{2}\\\\\therefore \ d=\pm \displaystyle \frac{1}{{\sqrt{2}}}\end{array}$ Since $ \displaystyle a<b<c$, $ \displaystyle {d=-\frac{1}{{\sqrt{2}}}}$ $ \displaystyle \begin{array}{l}\therefore \ \ a=\displaystyle \frac{1}{2}-\displaystyle \frac{1}{{\sqrt{2}}}\\\\\ \ \ \ b=\displaystyle \frac{1}{2}\\\\\ \ \ \ c=\displaystyle \frac{1}{2}+\displaystyle \frac{1}{{\sqrt{2}}}\end{array}$

8.         Find the sum to infinity of the series $ \displaystyle 1+\frac{2}{3}+\frac{4}{{{{3}^{2}}}}+\frac{{10}}{{{{3}^{3}}}}+\frac{{14}}{{{{3}^{4}}}}+...$

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Let $ \displaystyle S=1+\frac{2}{3}+\frac{4}{{{{3}^{2}}}}+\frac{{10}}{{{{3}^{3}}}}+\frac{{14}}{{{{3}^{4}}}}+... ---(1)$ $ \displaystyle \therefore \frac{1}{3}S=\frac{1}{3}+\frac{2}{{{{3}^{2}}}}+\frac{4}{{{{3}^{3}}}}+\frac{{10}}{{{{3}^{4}}}}+\frac{{14}}{{{{3}^{6}}}}+...\ ---(2)$ By equation $ \displaystyle (1)$ - equation $ \displaystyle (2)$ $ \displaystyle \begin{array}{l}\ \ \ \ S-\displaystyle \frac{1}{3}S=1+\displaystyle \frac{1}{3}+\displaystyle \frac{4}{{{{3}^{2}}}}+\displaystyle \frac{4}{{{{3}^{3}}}}+\displaystyle \frac{4}{{{{3}^{4}}}}+\displaystyle \frac{4}{{{{3}^{6}}}}+...\\\\\therefore \ \ S\left( {1-\displaystyle \frac{1}{3}} \right)=\displaystyle \frac{4}{3}+\displaystyle \frac{4}{{{{3}^{2}}}}\left( {1+\displaystyle \frac{1}{3}+\displaystyle \frac{1}{{{{3}^{2}}}}+\displaystyle \frac{1}{{{{3}^{3}}}}+...} \right)\\\\\therefore \ \ \displaystyle \frac{2}{3}S=\displaystyle \frac{4}{3}+\displaystyle \frac{4}{{{{3}^{2}}}}\left( {\displaystyle \frac{1}{{1-\displaystyle \frac{1}{3}}}} \right)\\\\\therefore \ \ \displaystyle \frac{2}{3}S=\displaystyle \frac{4}{3}+\displaystyle \frac{4}{{{{3}^{2}}}}\times \displaystyle \frac{3}{2}\\\\\therefore \ \ \displaystyle \frac{2}{3}S=2\\\\\therefore \,\ \ S=3\end{array}$

9.         If the $ \displaystyle (p+q)^{\text{th}}$ term of a $ \displaystyle G.P$ is $ \displaystyle m$ and the $ \displaystyle (p-q)^{\text{th}}$ term is $ \displaystyle n,$ find the $ \displaystyle p^{\text{th}}$ term in terms of $ \displaystyle m$ and $ \displaystyle n$.

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Let the first term of the $ \displaystyle G.P$ be $ \displaystyle a$ and the common ratio be $ \displaystyle r$. By the problem, $ \displaystyle \begin{array}{l}\,\ \ \ {{u}_{{p+q}}}=m\\\\\therefore \ \ a{{r}^{{p+q-1}}}=m\\\\\,\ \ \ {{u}_{{p-q}}}=n\\\\\therefore \ \ a{{r}^{{p-q-1}}}=n\\\\\therefore \ a{{r}^{{p+q-1}}}\times a{{r}^{{p-q-1}}}=mn\\\\\therefore \ {{a}^{2}}{{r}^{{2p-2}}}=mn\\\\\therefore \ {{(a{{r}^{{p-1}}})}^{2}}=mn\\\\\therefore \ a{{r}^{{p-1}}}=\sqrt{{mn}}\\\\\therefore \ {{u}_{p}}=\sqrt{{mn}}\end{array}$

10.      If $ \displaystyle x>1$ and $ \displaystyle {{\log }_{2}}x$, $ \displaystyle {{\log }_{3}}x$, $ \displaystyle {{\log }_{x}}16$ are in $ \displaystyle G.P.$, find the value of $ \displaystyle x$.

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$ \displaystyle {{\log }_{3}}x$, $ \displaystyle {{\log }_{x}}16$ are in $ \displaystyle G.P.$. $ \displaystyle \begin{array}{l}\therefore \ \ \ \ \displaystyle \frac{{{{{\log }}_{3}}x}}{{{{{\log }}_{2}}x}}=\displaystyle \frac{{{{{\log }}_{x}}16}}{{{{{\log }}_{3}}x}}\\\\\therefore \ \ \ \ {{\left( {{{{\log }}_{3}}x} \right)}^{2}}={{\log }_{2}}x\cdot {{\log }_{x}}16\\\\\therefore \ \ \ \ {{\left( {{{{\log }}_{3}}x} \right)}^{2}}=\displaystyle \frac{{{{{\log }}_{x}}16}}{{{{{\log }}_{x}}2}}\ \left[ {\because {{{\log }}_{b}}a=\displaystyle \frac{1}{{{{{\log }}_{a}}b}}} \right]\\\\\therefore \ \ \ \ {{\left( {{{{\log }}_{3}}x} \right)}^{2}}={{\log }_{2}}16\ \left[ {\because \displaystyle \frac{{{{{\log }}_{b}}x}}{{{{{\log }}_{b}}y}}={{{\log }}_{y}}x} \right]\\\\\therefore \ \ \ \ {{\left( {{{{\log }}_{3}}x} \right)}^{2}}={{\log }_{2}}{{2}^{4}}\\\\\therefore \ \ \ \ {{\left( {{{{\log }}_{3}}x} \right)}^{2}}=4\\\\\therefore \ \ \ \ {{\log }_{3}}x=\pm 2\\\\\therefore \ \ \ x={{3}^{2}}\ (\text{or})\ x={{3}^{{-2}}}\\\\\therefore \ \ \ x=9\ (\text{or})\ x=\displaystyle \frac{1}{9}\end{array}$ Since $ \displaystyle x>1$, $ \displaystyle x=\frac{1}{9}$ is impossible. $ \displaystyle \therefore \ \ \ x=9$

11.     How many terms of the sesies $ \displaystyle \sqrt{3}+ 3+ 3\sqrt{3}+ ...$ gives the sum $ \displaystyle 39+13\sqrt{3}$.

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Given series : $ \displaystyle \sqrt{3}+ 3+ 3\sqrt{3}+ ...$ $ \displaystyle \left. \begin{array}{l}\displaystyle \frac{3}{{\sqrt{3}}}=\sqrt{3}\\\\\displaystyle \frac{{3\sqrt{3}}}{3}=\sqrt{3}\end{array} \right\}\text{yields}\ \text{common ratio}$ $ \displaystyle \therefore\ \ \ \ $ Given terms are in $ \displaystyle G.P.$ with $ \displaystyle a=\sqrt{3}$ and $ \displaystyle d=\sqrt{3}$. By the problem, $ \displaystyle \begin{array}{l}\ \ \ \ {{S}_{n}}=39+13\sqrt{3}\\\\\therefore \ \ \displaystyle \frac{{a\left( {{{r}^{n}}-1} \right)}}{{r-1}}=39+13\sqrt{3}\\\\\therefore \ \ \displaystyle \frac{{\sqrt{3}\left( {{{{\left( {\sqrt{3}} \right)}}^{n}}-1} \right)}}{{\sqrt{3}-1}}=13\sqrt{3}\left( {\sqrt{3}+1} \right)\\\\\therefore \ \ \sqrt{3}\left( {{{{\left( {\sqrt{3}} \right)}}^{n}}-1} \right)=13\sqrt{3}\left( {\sqrt{3}+1} \right)\left( {\sqrt{3}-1} \right)\\\\\therefore \ \ {{\left( {\sqrt{3}} \right)}^{n}}-1=13\left( {3-1} \right)\\\\\therefore \ \ {{\left( {\sqrt{3}} \right)}^{n}}=27\\\\\therefore \ \ {{\left( {\sqrt{3}} \right)}^{n}}=\ {{\left( {\sqrt{3}} \right)}^{6}}\\\\\therefore \ \ n=6\end{array}$

Arithmetic Progression : Problems and Solutions

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

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Let $ \displaystyle A$ and $ \displaystyle d$ be the first term and the common difference respectively of the given $ \displaystyle A.P.,$ By the problem, $ \displaystyle \begin{array}{l}\ \ \ \ \ {{u}_{p}}=a\\\\\therefore \ \ \ A+(p-1)d=a\\\\\ \ \ \ \ {{u}_{q}}=b\\\\\therefore \ \ \ A+(q-1)d=b\\\\\ \ \ \ \ {{u}_{r}}=c\\\\\therefore \ \ \ A+(r-1)d=c\\\\\therefore \ \ \ a(q-r)\\\\\ \ =\ \left[ {A+(p-1)d} \right](q-r)\\\\\ \ =\ (q-r)A+(p-1)(q-r)d\\\\\ \ =\ (q-r)A+(pq-pr-q+r)d\\\\\ \ \ \ \ b(r-p)\\\\\ \ =\ \left[ {A+(q-1)d} \right](r-p)\\\\\ \ =\ (r-p)A+(qr-pq-r+p)d\\\\\ \ \ \ \ c(p-q)\\\\\ \ =\ \left[ {A+(r-1)d} \right](p-q)\\\\\ \ =\ (p-q)A+(r-1)(p-q)d\\\\\ \ =\ (p-q)A+(pr-qr-p+q)d\\\\\therefore \ \ \ a(q-r)+b(r-p)+c(p-q)\\\\\ \ =(q-r+r-p+p-q)A\\\ \ \ \ +(pq-pr-q+r+qr-pq-r+p+pr-qr-p+q)d\\\\\ \ =0\ \ \ \ \end{array}$

2.         Find the four numbers in $ \displaystyle A.P.$ such that their sum is $ \displaystyle 50$ and the greatest of them is four times the least.

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Let the four numbers in $ \displaystyle A.P.$ in ascending order be $ \displaystyle a$, $ \displaystyle a+d$, $ \displaystyle a+2d$ and $ \displaystyle a+3d$ respectively. $ \displaystyle \begin{array}{l}\therefore \ \ \ a+a+d+a+2d+a+3d=50\\\\\therefore \ \ \ 4a+6d=50\\\\\therefore \ \ \ 2a+3d=25\ ---(1)\\\\\therefore \ \ \ a+3d=4a\ \left[ {\text{given}} \right]\\\\\therefore \ \ \ 3a-3d=0---(2)\\\\\ \ \ \ \ \text{By}\ (1)+(2),\\\\\ \ \ \ \ 5a=25\\\\\therefore \ \ \ a=5\\\\\ \ \ \ \ \text{Substituting }a=5\ \text{in}\ \text{(1),}\\\\\ \ \ \ \ 2(5)+3d=25\\\\\therefore \ \ \ d=5\\\\\therefore \ \ \ {{1}^{{\text{st}}}}\ \text{number}=5\\\\\ \ \ \ \ {{2}^{{\text{nd}}}}\ \text{number}=10\\\\\ \ \ \ \ {{3}^{{\text{rd}}}}\ \text{number}=15\\\\\ \ \ \ \ {{4}^{{\text{th}}}}\ \text{number}=20\end{array}$

3.        Show that $ \displaystyle (a-b)^2$, $ \displaystyle (a^2+b^2)$ and $ \displaystyle (a+b)^2$ are in $ \displaystyle A.P.$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \,\ \left( {{{a}^{2}}+{{b}^{2}}} \right)-{{\left( {a-b} \right)}^{2}}\\\\\ \ \ \ \ =\left( {{{a}^{2}}+{{b}^{2}}} \right)-\left( {{{a}^{2}}-2ab+{{b}^{2}}} \right)\\\\\ \ \ \ \ ={{a}^{2}}+{{b}^{2}}-{{a}^{2}}+2ab-{{b}^{2}}\\\\\ \ \ \ \ =2ab\\\\\ \ \ \ \ \ \,\ {{\left( {a+b} \right)}^{2}}-\left( {{{a}^{2}}+{{b}^{2}}} \right)\\\\\ \ \ \ \ =\left( {{{a}^{2}}+2ab+{{b}^{2}}} \right)-\left( {{{a}^{2}}+{{b}^{2}}} \right)\\\\\ \ \ \ \ ={{a}^{2}}+2ab+{{b}^{2}}-{{a}^{2}}-{{b}^{2}}\\\\\ \ \ \ \ =2ab\\\\\therefore \ \ \ \left( {{{a}^{2}}+{{b}^{2}}} \right)-{{\left( {a-b} \right)}^{2}}={{\left( {a+b} \right)}^{2}}-\left( {{{a}^{2}}+{{b}^{2}}} \right)\end{array}$ $ \displaystyle \therefore$    $ \displaystyle (a^2+b^2)$ and $ \displaystyle (a+b)^2$ are in $ \displaystyle A.P.$

4.        If the sum of the first $ \displaystyle n$ terms of an $\displaystyle A.P$ is $ \displaystyle Pn+Qn^2$ where $ \displaystyle P$ and $ \displaystyle Q$ are real numbers, show that the common difference of that $\displaystyle A.P$ is $\displaystyle 2Q.$

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Let the common difference be $ \displaystyle d$, the $ \displaystyle n^{\text{th}}$ term be $ \displaystyle u_n$ and the sum of the first $ \displaystyle n$ terms be$ \displaystyle S_n$ of given $ \displaystyle A.P.$ $ \displaystyle \begin{array}{l}\therefore \ \ \ \ \ \ {{S}_{n}}=Pn+Q{{n}^{2}}\\\\\therefore \ \ \ \ \ \ {{S}_{{n-1}}}=P\left( {n-1} \right)+Q{{\left( {n-1} \right)}^{2}}\\\\\,\ \ \ \ \ \ \ \ \ \ \ \ \ \ =Pn-P+Q{{n}^{2}}-2Qn+Q\\\\\ \ \ \ \ \ \ \ \text{Since}\ {{u}_{n}}={{S}_{n}}-{{S}_{{n-1}}},\\\\\ \ \ \ \ \ \ \ {{u}_{n}}=2Qn-P-Q\\\\\therefore \ \ \ \ \ \ {{u}_{{n-1}}}=2Q\left( {n-1} \right)-P-Q\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =2Qn-P-3Q\\\\\ \ \ \ \ \ \ \ \text{Since}\ d={{u}_{n}}-{{u}_{{n-1}}},\\\\\ \ \ \ \ \ \ \ d=2Q\end{array}$

5.        If $ \displaystyle A$ is the single arithmetic mean and $ \displaystyle S$ is $ \displaystyle n$ arithmetic means between $ \displaystyle a$ and $ \displaystyle b$, show that $ \displaystyle \frac{S}{A}=n$.

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Let $ \displaystyle a, a+d,$ ..., $ \displaystyle b-d, b$ be an $ \displaystyle A.P.$ of $ \displaystyle n+2$ terms where $ \displaystyle d$ be a common difference. $ \displaystyle \therefore \ \ \ \ \ A=\frac{{a+b}}{2}$ $ \displaystyle a+d,$ ..., $ \displaystyle b-d$ are $ \displaystyle n$ arithmetic means between $ \displaystyle a$ and $ \displaystyle b$. $ \displaystyle \begin{array}{l}\therefore \ \ \ \ \ S=\displaystyle \frac{n}{2}\left( {a+d+b-d} \right)\\\ \ \ \ \ \ \ \left[ {\because {{S}_{n}}=\displaystyle \frac{n}{2}(a+l)} \right]\\\\\therefore \ \ \ \ S=\displaystyle \frac{n}{2}\left( {a+b} \right)\\\\\therefore \ \ \ \ \displaystyle \frac{S}{A}=\displaystyle \frac{{\displaystyle \frac{n}{2}\left( {a+b} \right)}}{{\displaystyle \frac{{a+b}}{2}}}\\\\\therefore \ \ \ \ \displaystyle \frac{S}{A}=n\end{array}$

6.        If, in an $ \displaystyle A.P.$, $ \displaystyle S_n={n}^{2}p$ and $ \displaystyle S_m={m}^{2}p$ where $ \displaystyle m \ne n$, then prove that $ \displaystyle S_p={p}^{3}$.

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Let $ \displaystyle a$ be the first term and Let $ \displaystyle d$ be the common difference of given $ \displaystyle A.P.$        By the problem, $ \displaystyle \begin{array}{l}\ \ \ \ \ \ {{S}_{n}}={{n}^{2}}p\\\\\therefore \ \ \ \ \displaystyle \frac{n}{2}\{2a+(n-)d\}={{n}^{2}}p\\\\\therefore \ \ \ \ 2a+(n-1)d=2np ---(1)\end{array}$        Similarly, $ \displaystyle \begin{array}{l}\ \ \ \ \ \ {{S}_{m}}={{m}^{2}}p\\\\\therefore \ \ \ \ \displaystyle \frac{m}{2}\{2a+(m-)d\}={{m}^{2}}p\\\\\therefore \ \ \ \ 2a+(m-1)d=2mp ---(2)\end{array}$        By $ \displaystyle (1)-(2)$, $ \displaystyle \ \ \ \ \ \ (n-m)d=2p(n-m)$        Since $ \displaystyle m \ne n, n-m\ne 0$, $ \displaystyle \begin{array}{l}\therefore \ \ \ \ d=2p\\\\\therefore \ \ \ \ 2a+(n-1)(2p)=2np\\\\\therefore \ \ \ \ 2a+2np-2p=2np\\\\\therefore \ \ \ \ a=p\\\\\therefore \ \ \ \ {{S}_{p}}=\displaystyle \frac{p}{2}\left\{ {2a+(p-1)d} \right\}\\\\\ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{p}{2}\left\{ {2p+(p-1)(2p)} \right\}\\\\\ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{p}{2}(2p)\left\{ {1+p-1} \right\}\\\\\ \ \ \ \ \ \ \ \ \ \ ={{p}^{3}}\end{array}$

7.       A certain $ \displaystyle A.P.$ has even number of terms. If the sum of odd terms is $ \displaystyle24,$ the sum of even terms is $ \displaystyle 30$ and the last term is $ \displaystyle 10\frac{1}{2}$ more than the first term, find the number of terms in that $ \displaystyle A.P.$

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Let the first term be $ \displaystyle a$, the common be $ \displaystyle d$ and the number of terms contains in that $ \displaystyle A.P.$ be $ \displaystyle n$. Since given $ \displaystyle A.P.$ contains even number of terms, assume that $ \displaystyle n=2m$. Let the given $ \displaystyle A.P.$ be $ \displaystyle a, a+d, ..., a+(2m-1)d$. By the problem, $ \displaystyle \ \ \ \ \ \ a+\left(a+2d\right)+ ... + \left(a+2md\right)=24$. $ \displaystyle \therefore \ \ \ \frac{m}{2}\left[ {a+a+(2m-2)d} \right]=24\ \ $ စံု ႀကိမ္ေျမာက္ကိန္းနဲ႔ မ ႀကိမ္ေျမာက္ကိန္း တစ္၀က္စီရွိပါတယ္။ စုစုေပါင္း ကိန္းလံုး အေရအတြက္က $ \displaystyle 2m$ ျဖစ္လို႔ စံုႀကိမ္ေျမာက္ ကိန္းအေရအတြက္၊ မ ႀကိမ္ေျမာက္ ကိန္း အေရအတြက္၊ ႏွစ္ခုလံုးက $ \displaystyle m$ ျဖစ္တယ္လို႔ သိရမယ္။ $ \displaystyle \begin{array}{l}\therefore \ \ \ \displaystyle \frac{m}{2}\left[ {2a+2(m-1)d} \right]=24\\\\\therefore \ \ \ ma+{{m}^{2}}d-md=24---(1)\end{array}$ Again, $ \displaystyle \left(a+d\right)+\left(a+3d\right)+ ... + \left[a+(2m-1)d\right]=30$. $ \displaystyle \begin{array}{l}\therefore \ \ \ \displaystyle \frac{m}{2}\left[ {a+d+a+(2m-1)d} \right]=30\\\\\therefore \ \ \ \displaystyle \frac{m}{2}\left[ {2a+2md} \right]=30\\\\\therefore \ \ \ ma+{{m}^{2}}d=30---(2)\end{array}$ By $ \displaystyle (2)-(1)$, $ \displaystyle \begin{array}{l}\ \ \ \ \ md=6\\\\\therefore \ \ \ d=\displaystyle \frac{6}{m}\ \ \end{array}$ By the problem, last term = first term + $ \displaystyle 10\frac{1}{2}$ $ \displaystyle \begin{array}{l}\therefore \ \ \ a+(2m-1)d=a+10\displaystyle \frac{1}{2}\\\\\therefore \ \ \ \displaystyle \frac{{24}}{m}(2m-1)=\displaystyle \frac{{21}}{2}\\\\\therefore \ \ \ \displaystyle \frac{6}{m}(2m-1)=\displaystyle \frac{{21}}{2}\\\\\therefore \ \ \ 4(2m-1)=7m\\\\\therefore \ \ \ m=4\\\\\therefore \ \ \ n=8\end{array}$

8.        If $ \displaystyle S_n$ denotes sum of $ \displaystyle n$ terms of an $ \displaystyle A.P.$ and if $ \displaystyle S_1= 6, S_7 = 105$, then prove that $ \displaystyle \frac{{{{S}_{n}}}}{{{{S}_{{n-3}}}}}=\frac{{n+3}}{{n-3}}$.

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Let the first term be $ \displaystyle a$ and the common be $ \displaystyle d$ of the given $ \displaystyle A.P.$ $ \displaystyle \therefore \ \ {{S}_{n}}=\frac{n}{2}\left\{ {2a+\left( {n-1} \right)d} \right\}$ By the problem, $ \displaystyle \begin{array}{l}\ \ \ {{S}_{1}}=6\\\\\therefore \ \ a=6\\\\\ \ \ {{S}_{7}}=105\\\\\therefore \ \ \displaystyle \frac{7}{2}\left\{ {12+6d} \right\}=105\\\\\therefore \ \ d=3\\\\\therefore \ \displaystyle \frac{{{{S}_{n}}}}{{{{S}_{{n-3}}}}}=\displaystyle \frac{{\displaystyle \frac{n}{2}\left\{ {2a+\left( {n-1} \right)d} \right\}}}{{\displaystyle \frac{{n-3}}{2}\left\{ {2a+\left( {n-3-1} \right)d} \right\}}}\\\\\therefore \ \displaystyle \frac{{{{S}_{n}}}}{{{{S}_{{n-3}}}}}=\displaystyle \frac{{n\left\{ {12+\left( {n-1} \right)3} \right\}}}{{\left( {n-3} \right)\left\{ {12+\left( {n-3-1} \right)3} \right\}}}\\\\\therefore \ \displaystyle \frac{{{{S}_{n}}}}{{{{S}_{{n-3}}}}}=\displaystyle \frac{{3n(n+3)}}{{3n\left( {n-3} \right)}}\\\\\therefore \ \displaystyle \frac{{{{S}_{n}}}}{{{{S}_{{n-3}}}}}=\displaystyle \frac{{n+3}}{{n-3}}\end{array}$

9.        If the $ \displaystyle A.M.$ between $ \displaystyle {p}^{\text{th}}$ and $ \displaystyle {q}^{\text{th}}$ terms of an $ \displaystyle A.P.$ is equal to the $ \displaystyle A.M.$ between $ \displaystyle {r}^{\text{th}}$ and $ \displaystyle {s}^{\text{th}}$ terms of the $ \displaystyle A.P.$, then show that $ \displaystyle (p + q) = (r + s)$.

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Let the $ \displaystyle {1}^{\text{st}}$ be $ \displaystyle a$ and the common difference be $ \displaystyle d$ of given $ \displaystyle A.P.$ By the problem, $ \displaystyle A.M.$ between $ \displaystyle u_p$ and $ \displaystyle u_q$ =$ \displaystyle A.M.$ between $ \displaystyle u_r$ and $ \displaystyle u_s$ $ \displaystyle \begin{array}{l}\therefore \ \ \displaystyle \frac{{{{u}_{p}}+{{u}_{q}}}}{2}=\ \displaystyle \frac{{{{u}_{r}}+{{u}_{s}}}}{2}\\\\\therefore \ \ {{u}_{p}}+{{u}_{q}}=\ {{u}_{r}}+{{u}_{s}}\\\\\therefore \ \ a+(p-1)d+a+(q-1)d=\ a+(r-1)d+a+(s-1)d\\\\\therefore \ \ (p+q-2)d=\ (r+s-2)d\\\\\therefore \ \ p+q-2=\ r+s-2\\\\\therefore \ \ p+q=\ r+s\end{array}$

10.     The sum to $ \displaystyle n$ terms of an $ \displaystyle A.P.$ is $ \displaystyle 3n^2+5n$ and the $ \displaystyle {p}^{\text{th}}$ term of that $ \displaystyle A.P.$ is $ \displaystyle 164.$ Find the value of $ \displaystyle p.$

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By the problem, $ \displaystyle \begin{array}{l} {{S}_{n}}=3{{n}^{2}}+5n\\\\ {{u}_{p}}=164\end{array}$ Since $ \displaystyle {{u}_{p}}={{S}_{p}}-{{S}_{{p-1}}}.$ $ \displaystyle \begin{array}{l}164=\left[ {3{{p}^{2}}+5p} \right]-\left[ {3{{{\left( {p-1} \right)}}^{2}}+5\left( {p-1} \right)} \right]\\\\164=\left[ {3{{p}^{2}}+5p} \right]-\left[ {3{{p}^{2}}-6p+3+5p-5} \right]\\\\164=6p+2\\\\\therefore \ 6p=162\\\\\therefore \ p=27\ \ \ \end{array}$