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Question Number 73247 by Lontum Hans last updated on 09/Nov/19

Answered by Kunal12588 last updated on 09/Nov/19

a_8 −a_4 =40 ....(1) a_8 =(3/2)a_4 .....(2) put a_8 in eq^n (1) (3/2)a_4 −a_4 =40 ⇒(a_4 /2)=40 ⇒a_4 =80 put a_4 in eq^n (2) a_8 =(3/2)×80 ⇒a_8 =120 a_4 =80 ⇒a_1 +3d=80 ....(3) a_8 =120 ⇒a_1 +7d=120 ....(4) eq^n (4)−eq^n (3) 4d=40 ⇒d=10 put d in eq^n (3) a_1 +30=80 ⇒a_1 =50 ans(1) first term, i.e. a_1 =50 ans(2) common difference, i.e. d=10 ans(3) a_9 =a_1 +8d=50+80=130 a_(10) =a_9 +d=130+10=140 a_(21) =a_1 +20d=50+200=250 a_(32) =a_1 +31d=50+310=360 a_(41) =a_1 +40d=50+400=450 ans(4) S_n =(n/2)×{2a+(n−1)d} S_4 =(4/2)×{2×50+3×10}=2(100+30)=260 S_5 =(5/2)×{2×50+4×10}=(5/2)(100+40)=350

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

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