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.