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\ \ $
$ \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}$ |
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