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