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Question-210630




Question Number 210630 by peter frank last updated on 14/Aug/24
Answered by mm1342 last updated on 14/Aug/24
a_n =a+(n−1)d   &  b_n =2^(n−1) g  c_n =a_n +b_n  &  c_1 =57  &  c_3 =171  & c_5 =645  ⇒a+g=57    &   a+2d+4g=171  &  a+4d+16d=645  ⇒a=17  &  d=−3  &  g=40  ⇒c_n =40×2^(n−1) −3n+20 ✓
$${a}_{{n}} ={a}+\left({n}−\mathrm{1}\right){d}\:\:\:\&\:\:{b}_{{n}} =\mathrm{2}^{{n}−\mathrm{1}} {g} \\ $$$${c}_{{n}} ={a}_{{n}} +{b}_{{n}} \:\&\:\:{c}_{\mathrm{1}} =\mathrm{57}\:\:\&\:\:{c}_{\mathrm{3}} =\mathrm{171}\:\:\&\:{c}_{\mathrm{5}} =\mathrm{645} \\ $$$$\Rightarrow{a}+{g}=\mathrm{57}\:\:\:\:\&\:\:\:{a}+\mathrm{2}{d}+\mathrm{4}{g}=\mathrm{171}\:\:\&\:\:{a}+\mathrm{4}{d}+\mathrm{16}{d}=\mathrm{645} \\ $$$$\Rightarrow{a}=\mathrm{17}\:\:\&\:\:{d}=−\mathrm{3}\:\:\&\:\:{g}=\mathrm{40} \\ $$$$\Rightarrow{c}_{{n}} =\mathrm{40}×\mathrm{2}^{{n}−\mathrm{1}} −\mathrm{3}{n}+\mathrm{20}\:\checkmark \\ $$$$\:\:\: \\ $$

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