Question Number 210630 by peter frank last updated on 14/Aug/24
Answered by mm1342 last updated on 14/Aug/24
$${a}_{{n}} ={a}+\left({n}−\mathrm{1}\right){d}\:\:\:\&\:\:{b}_{{n}} =\mathrm{2}^{{n}−\mathrm{1}} {g} \\ $$$${c}_{{n}} ={a}_{{n}} +{b}_{{n}} \:\&\:\:{c}_{\mathrm{1}} =\mathrm{57}\:\:\&\:\:{c}_{\mathrm{3}} =\mathrm{171}\:\:\&\:{c}_{\mathrm{5}} =\mathrm{645} \\ $$$$\Rightarrow{a}+{g}=\mathrm{57}\:\:\:\:\&\:\:\:{a}+\mathrm{2}{d}+\mathrm{4}{g}=\mathrm{171}\:\:\&\:\:{a}+\mathrm{4}{d}+\mathrm{16}{d}=\mathrm{645} \\ $$$$\Rightarrow{a}=\mathrm{17}\:\:\&\:\:{d}=−\mathrm{3}\:\:\&\:\:{g}=\mathrm{40} \\ $$$$\Rightarrow{c}_{{n}} =\mathrm{40}×\mathrm{2}^{{n}−\mathrm{1}} −\mathrm{3}{n}+\mathrm{20}\:\checkmark \\ $$$$\:\:\: \\ $$