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Daca suma [tex] S_{n} [/tex] a prmilor n termeni ai unei progresi aritmetice este [tex] 8n^{2} -6n[/tex] atunci ratia este?
a)2 b)16 c)18 d)20

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Răspuns :

metoda I
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notez a1,a2,...,an cei n termeni si r,ratia.
a1=a1
a2=a1+r
a3=a2+r=a1+2*r
................
an=a1+(n-1)*r

=> Sn=a1+a2+...+an =a1+(a1+r)+...+(a1+ (n-1) )=
         =n*a1 + r*(1+2+...+ (n-1) )=
         =n*a1+ r*n*(n-1)/2
dar Sn=8*n^2-6*n =>
=> n*a1+ r*n*(n-1)/2 =8*n^2-6*n <=>
<=>2*a1*n+ r*n^2-r*n =16*n^2-12*n<=>
<=>r*n^2 -n(r-2*a1) =16*n^2-12n <=>
<=>r=16    si r-2*a1=12 => 16-2*a1=12 =>2*a1=5=>a1=2
raspuns b)

metoda II
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daca Sn=8*n^2-6*n
atunci S(1)=8*1^2-6*1=8-6=2
        S(2)=8*2^2-6*2=32-12=20
       
cum    S(1)=a1 =2
iar       S(2) -S(1) =a1+a2 -a1 =a2     =20-2 =18
si a2=a1+r    => 18=2+r => r=16