Menu Close

Question-21504




Question Number 21504 by mondodotto@gmail.com last updated on 25/Sep/17
Answered by $@ty@m last updated on 26/Sep/17
ATQ  S_n =(−1)+1+3+5+......+(2n−3)  (i) S_(50) =((50)/2){(−1)+(2×50−3)}  =25(−1+97)  =25×96  =2400  (ii) S_n =624  (n/2){2a+(n−1)d}=624  (n/2){−2+(n−1)×2}=624  n(−1+n−1)=624  n(n−2)=624  n^2 −2n−624=0  n=((2±(√(4+2496)))/2)  n=((2±50)/2)  n=26,−24  n=26 {∵n∈N}
$${ATQ} \\ $$$$\boldsymbol{{S}}_{{n}} =\left(−\mathrm{1}\right)+\mathrm{1}+\mathrm{3}+\mathrm{5}+……+\left(\mathrm{2}{n}−\mathrm{3}\right) \\ $$$$\left({i}\right)\:\boldsymbol{{S}}_{\mathrm{50}} =\frac{\mathrm{50}}{\mathrm{2}}\left\{\left(−\mathrm{1}\right)+\left(\mathrm{2}×\mathrm{50}−\mathrm{3}\right)\right\} \\ $$$$=\mathrm{25}\left(−\mathrm{1}+\mathrm{97}\right) \\ $$$$=\mathrm{25}×\mathrm{96} \\ $$$$=\mathrm{2400} \\ $$$$\left({ii}\right)\:\boldsymbol{{S}}_{{n}} =\mathrm{624} \\ $$$$\frac{{n}}{\mathrm{2}}\left\{\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right\}=\mathrm{624} \\ $$$$\frac{{n}}{\mathrm{2}}\left\{−\mathrm{2}+\left({n}−\mathrm{1}\right)×\mathrm{2}\right\}=\mathrm{624} \\ $$$${n}\left(−\mathrm{1}+{n}−\mathrm{1}\right)=\mathrm{624} \\ $$$${n}\left({n}−\mathrm{2}\right)=\mathrm{624} \\ $$$${n}^{\mathrm{2}} −\mathrm{2}{n}−\mathrm{624}=\mathrm{0} \\ $$$${n}=\frac{\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{2496}}}{\mathrm{2}} \\ $$$${n}=\frac{\mathrm{2}\pm\mathrm{50}}{\mathrm{2}} \\ $$$${n}=\mathrm{26},−\mathrm{24} \\ $$$${n}=\mathrm{26}\:\left\{\because{n}\in\mathrm{N}\right\} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *