Question

determinati primul termen si ratia unei progresii aritmetice stiind ca s6=141 si ca s8=244

Answer 1

Răspuns:

a₁ = 6 ; r = 7

Explicație pas cu pas:

S₆ = 141

S₈ = 244

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S₆ = a₁+a₂+.....+a₆ = a₁+a₁+r+a₁+2r+...+a₁+5r =

= 6a₁+(1+2+3+4+5)·r = 6a₁+15r = 141

S₈ = S₆+a₇+a₈ = 6a₁+15r+a₁+6r+a₁+7r = 8a₁+28r = 244

6a₁+15r = 141   <=> 2a₁+5r = 47

8a₁+28r = 244 <=> 2a₁+7r = 61  =>

7r-5r = 61-47 => 2r = 14 => r = 7

2a₁ +35 = 47 => 2a₁ = 47-35 = 12 => a₁ = 6

Sirul progresiei aritmetice :

6 ; 13 ; 20 ; 27 ; 34 ; 41 ; 48 ; 55 ; 62 ........

S₆ = 6+13+20+27+34+41 = 141

S₈ = 6+13+20+27+34+41+48+55 = 244