1. If the pth, qth and rth terms of an A.P. are a,b, and c respectively, prove that a(q−r) + b(r−p) + c(p−q)=0.
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| Let A and d be the first term and the common difference respectively of the given A.P., By the problem, up=a∴ |
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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