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Question Number 41381 by behi83417@gmail.com last updated on 06/Aug/18

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Aug/18

question no 2) is it AP or GP

$$\left.{question}\:{no}\:\mathrm{2}\right)\:{is}\:{it}\:{AP}\:{or}\:{GP} \\ $$

Commented by behi83417@gmail.com last updated on 06/Aug/18

2# given AP and not AP in#1.

$$\mathrm{2}#\:{given}\:{AP}\:{and}\:{not}\:{AP}\:{in}#\mathrm{1}. \\ $$

Commented by $@ty@m last updated on 06/Aug/18

AP has a common difference not a common ratio.

$${AP}\:{has}\:{a}\:{common}\:{difference} \\ $$$${not}\:{a}\:{common}\:{ratio}. \\ $$

Commented by behi83417@gmail.com last updated on 06/Aug/18

ok.you are right.it is a typo. sum of n terms=N sum of m terms=M common difference=?

$${ok}.{you}\:{are}\:{right}.{it}\:{is}\:{a}\:{typo}. \\ $$$$\boldsymbol{{sum}}\:\boldsymbol{{of}}\:\boldsymbol{{n}}\:\boldsymbol{{terms}}=\boldsymbol{{N}} \\ $$$$\boldsymbol{{sum}}\:\boldsymbol{{of}}\:\boldsymbol{{m}}\:\boldsymbol{{terms}}=\boldsymbol{{M}} \\ $$$$\boldsymbol{{common}}\:\boldsymbol{{difference}}=? \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Aug/18

A+(m−1)D=(1/n) A+(n−1)D=(1/m) (m−n)D=(1/n)−(1/m) (m−n)D=((m−n)/(mn)) D=(1/(mn)) A+(m−1)×(1/(mn))=(1/n) A=(1/n)−((m−1)/(mn))=((m−m+1)/(mn))=(1/(mn)) so a_(mn) =A+(mn−1)×D =(1/(mn))+(mn−1)×(1/(mn)) =(1/(mn))(1+mn−1)=1

$${A}+\left({m}−\mathrm{1}\right){D}=\frac{\mathrm{1}}{{n}} \\ $$$${A}+\left({n}−\mathrm{1}\right){D}=\frac{\mathrm{1}}{{m}} \\ $$$$\left({m}−{n}\right){D}=\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{m}} \\ $$$$\left({m}−{n}\right){D}=\frac{{m}−{n}}{{mn}}\:\:\:\:{D}=\frac{\mathrm{1}}{{mn}} \\ $$$${A}+\left({m}−\mathrm{1}\right)×\frac{\mathrm{1}}{{mn}}=\frac{\mathrm{1}}{{n}} \\ $$$${A}=\frac{\mathrm{1}}{{n}}−\frac{{m}−\mathrm{1}}{{mn}}=\frac{{m}−{m}+\mathrm{1}}{{mn}}=\frac{\mathrm{1}}{{mn}} \\ $$$${so}\:{a}_{{mn}} ={A}+\left({mn}−\mathrm{1}\right)×{D} \\ $$$$=\frac{\mathrm{1}}{{mn}}+\left({mn}−\mathrm{1}\right)×\frac{\mathrm{1}}{{mn}} \\ $$$$=\frac{\mathrm{1}}{{mn}}\left(\mathrm{1}+{mn}−\mathrm{1}\right)=\mathrm{1} \\ $$$$ \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Aug/18

2)N=(n/2)[2a+(n−1)d]] M=(m/2).[2a+(n−1d] 2a+(n−1)d=((2N)/n) 2a+(m−1)d=((2M)/m) (n−1−m+1)d=((2N)/n)−((2M)/m) d=((2((N/n)−(M/m)))/(n−m))

$$\left.\mathrm{2}\left.\right){N}=\frac{{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right]\right] \\ $$$$\:\:{M}=\frac{{m}}{\mathrm{2}}.\left[\mathrm{2}{a}+\left({n}−\mathrm{1}{d}\right]\right. \\ $$$$\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}=\frac{\mathrm{2}{N}}{{n}} \\ $$$$\mathrm{2}{a}+\left({m}−\mathrm{1}\right){d}=\frac{\mathrm{2}{M}}{{m}} \\ $$$$\left({n}−\mathrm{1}−{m}+\mathrm{1}\right){d}=\frac{\mathrm{2}{N}}{{n}}−\frac{\mathrm{2}{M}}{{m}} \\ $$$${d}=\frac{\mathrm{2}\left(\frac{{N}}{{n}}−\frac{{M}}{{m}}\right)}{{n}−{m}} \\ $$

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