Question Number 73247 by Lontum Hans last updated on 09/Nov/19
Answered by Kunal12588 last updated on 09/Nov/19
$${a}_{\mathrm{8}} −{a}_{\mathrm{4}} =\mathrm{40}\:\:\:….\left(\mathrm{1}\right) \\ $$$${a}_{\mathrm{8}} =\frac{\mathrm{3}}{\mathrm{2}}{a}_{\mathrm{4}} \:\:\:\:\:\:\:…..\left(\mathrm{2}\right) \\ $$$${put}\:{a}_{\mathrm{8}} \:{in}\:{eq}^{{n}} \left(\mathrm{1}\right) \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}{a}_{\mathrm{4}} −{a}_{\mathrm{4}} =\mathrm{40} \\ $$$$\Rightarrow\frac{{a}_{\mathrm{4}} }{\mathrm{2}}=\mathrm{40} \\ $$$$\Rightarrow{a}_{\mathrm{4}} =\mathrm{80} \\ $$$${put}\:{a}_{\mathrm{4}} \:{in}\:{eq}^{{n}} \left(\mathrm{2}\right) \\ $$$${a}_{\mathrm{8}} =\frac{\mathrm{3}}{\mathrm{2}}×\mathrm{80} \\ $$$$\Rightarrow{a}_{\mathrm{8}} =\mathrm{120} \\ $$$${a}_{\mathrm{4}} =\mathrm{80} \\ $$$$\Rightarrow{a}_{\mathrm{1}} +\mathrm{3}{d}=\mathrm{80}\:\:\:….\left(\mathrm{3}\right) \\ $$$${a}_{\mathrm{8}} =\mathrm{120} \\ $$$$\Rightarrow{a}_{\mathrm{1}} +\mathrm{7}{d}=\mathrm{120}\:\:….\left(\mathrm{4}\right) \\ $$$${eq}^{{n}} \left(\mathrm{4}\right)−{eq}^{{n}} \left(\mathrm{3}\right) \\ $$$$\mathrm{4}{d}=\mathrm{40} \\ $$$$\Rightarrow{d}=\mathrm{10} \\ $$$${put}\:{d}\:{in}\:{eq}^{{n}} \left(\mathrm{3}\right) \\ $$$${a}_{\mathrm{1}} +\mathrm{30}=\mathrm{80} \\ $$$$\Rightarrow{a}_{\mathrm{1}} =\mathrm{50} \\ $$$${ans}\left(\mathrm{1}\right)\:{first}\:{term},\:{i}.{e}.\:{a}_{\mathrm{1}} =\mathrm{50} \\ $$$${ans}\left(\mathrm{2}\right)\:{common}\:{difference},\:{i}.{e}.\:{d}=\mathrm{10} \\ $$$${ans}\left(\mathrm{3}\right) \\ $$$${a}_{\mathrm{9}} ={a}_{\mathrm{1}} +\mathrm{8}{d}=\mathrm{50}+\mathrm{80}=\mathrm{130} \\ $$$${a}_{\mathrm{10}} ={a}_{\mathrm{9}} +{d}=\mathrm{130}+\mathrm{10}=\mathrm{140} \\ $$$${a}_{\mathrm{21}} ={a}_{\mathrm{1}} +\mathrm{20}{d}=\mathrm{50}+\mathrm{200}=\mathrm{250} \\ $$$${a}_{\mathrm{32}} ={a}_{\mathrm{1}} +\mathrm{31}{d}=\mathrm{50}+\mathrm{310}=\mathrm{360} \\ $$$${a}_{\mathrm{41}} ={a}_{\mathrm{1}} +\mathrm{40}{d}=\mathrm{50}+\mathrm{400}=\mathrm{450} \\ $$$${ans}\left(\mathrm{4}\right) \\ $$$${S}_{{n}} =\frac{{n}}{\mathrm{2}}×\left\{\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right\} \\ $$$${S}_{\mathrm{4}} =\frac{\mathrm{4}}{\mathrm{2}}×\left\{\mathrm{2}×\mathrm{50}+\mathrm{3}×\mathrm{10}\right\}=\mathrm{2}\left(\mathrm{100}+\mathrm{30}\right)=\mathrm{260} \\ $$$${S}_{\mathrm{5}} =\frac{\mathrm{5}}{\mathrm{2}}×\left\{\mathrm{2}×\mathrm{50}+\mathrm{4}×\mathrm{10}\right\}=\frac{\mathrm{5}}{\mathrm{2}}\left(\mathrm{100}+\mathrm{40}\right)=\mathrm{350} \\ $$