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Question-190230




Question Number 190230 by 073 last updated on 29/Mar/23
Answered by Rasheed.Sindhi last updated on 29/Mar/23
a_5 =5−1=4   [ ∵  5=2(2)+1 ]  a_6 =2(6)+3=15   [ ∵ 6=2(3) ]  a_5 +a_6 =4+15=19
$$\mathrm{a}_{\mathrm{5}} =\mathrm{5}−\mathrm{1}=\mathrm{4}\:\:\:\left[\:\because\:\:\mathrm{5}=\mathrm{2}\left(\mathrm{2}\right)+\mathrm{1}\:\right] \\ $$$$\mathrm{a}_{\mathrm{6}} =\mathrm{2}\left(\mathrm{6}\right)+\mathrm{3}=\mathrm{15}\:\:\:\left[\:\because\:\mathrm{6}=\mathrm{2}\left(\mathrm{3}\right)\:\right] \\ $$$$\mathrm{a}_{\mathrm{5}} +\mathrm{a}_{\mathrm{6}} =\mathrm{4}+\mathrm{15}=\mathrm{19} \\ $$
Answered by a.lgnaoui last updated on 29/Mar/23
n=5 ⇒5 impair(=2×2+1)  a_5 =5−1=4  n=6⇒6 pair(6=2×3)  a_6 =2×6+3=15  alors:   a_5 +a_6 =15+4            a_5 +a_6 =19
$${n}=\mathrm{5}\:\Rightarrow\mathrm{5}\:{impair}\left(=\mathrm{2}×\mathrm{2}+\mathrm{1}\right) \\ $$$${a}_{\mathrm{5}} =\mathrm{5}−\mathrm{1}=\mathrm{4} \\ $$$${n}=\mathrm{6}\Rightarrow\mathrm{6}\:{pair}\left(\mathrm{6}=\mathrm{2}×\mathrm{3}\right) \\ $$$${a}_{\mathrm{6}} =\mathrm{2}×\mathrm{6}+\mathrm{3}=\mathrm{15} \\ $$$${alors}:\:\:\:{a}_{\mathrm{5}} +{a}_{\mathrm{6}} =\mathrm{15}+\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{{a}}_{\mathrm{5}} +\boldsymbol{{a}}_{\mathrm{6}} =\mathrm{19} \\ $$

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