Question Number 189183 by pascal889 last updated on 12/Mar/23
Answered by Rasheed.Sindhi last updated on 13/Mar/23
![a: first term of AP, g: first term of GP a_1 +g_1 =a+g=8⇒a=8−g a_3 =a+(3−1)d=a+2(2) g_3 =gr^(3−1) =g(3)^2 a_3 +g_3 =[a+2(2)]+[g(3)^2 ]=52 ⇒a+9g=48 ⇒8−g+9g=48⇒g=5⇒a=3 (i)AP: 3,5,7,... (ii)GP: 5,15,45,... (iii) The new sequence:8,20,52,...](Q189197.png)
$${a}:\:{first}\:{term}\:{of}\:{AP}, \\ $$$${g}:\:{first}\:{term}\:{of}\:{GP} \\ $$$${a}_{\mathrm{1}} +{g}_{\mathrm{1}} ={a}+{g}=\mathrm{8}\Rightarrow{a}=\mathrm{8}−{g} \\ $$$${a}_{\mathrm{3}} ={a}+\left(\mathrm{3}−\mathrm{1}\right){d}={a}+\mathrm{2}\left(\mathrm{2}\right) \\ $$$${g}_{\mathrm{3}} ={gr}^{\mathrm{3}−\mathrm{1}} ={g}\left(\mathrm{3}\right)^{\mathrm{2}} \\ $$$${a}_{\mathrm{3}} +{g}_{\mathrm{3}} =\left[{a}+\mathrm{2}\left(\mathrm{2}\right)\right]+\left[{g}\left(\mathrm{3}\right)^{\mathrm{2}} \right]=\mathrm{52} \\ $$$$\Rightarrow{a}+\mathrm{9}{g}=\mathrm{48} \\ $$$$\Rightarrow\mathrm{8}−{g}+\mathrm{9}{g}=\mathrm{48}\Rightarrow{g}=\mathrm{5}\Rightarrow{a}=\mathrm{3} \\ $$$$\left({i}\right){AP}:\:\mathrm{3},\mathrm{5},\mathrm{7},... \\ $$$$\left({ii}\right){GP}:\:\mathrm{5},\mathrm{15},\mathrm{45},... \\ $$$$\left({iii}\right)\:{The}\:{new}\:{sequence}:\mathrm{8},\mathrm{20},\mathrm{52},... \\ $$
Commented by pascal889 last updated on 13/Mar/23
$${please}\:{i}\:{dont}\:{understand}\:{the}\:{second}\:{line}\:{sir} \\ $$
Commented by Rasheed.Sindhi last updated on 13/Mar/23
$${I}'{ve}\:{added}\:{some}\:{lines}\:{to}\:{describe}. \\ $$