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2009-11-13T20:09:55+01:00
A) a3 = 14
S3 - 30
a3 = a1 + 2r = 14
a1 = 14 -2r oraz S3 = a1 +a2 +a3 = a1 +(a1 +r)+(a1 +2r)
S3 = 3a1 + 3r = 3*( 14 - 2r) + 3r = 42 - 6r +3r =42 -3r =30
3r =42 - 30 - 12
r = 12 : 3 = 4
r = 4
a1 = 14 - 2r = 14 -2*4 = 14 - 8 = 6
an =a1 +(n-1)*r = 6 +(n-1)*4 = 6 -4 +4n = 4n +2

c)
a5 = 15
S4 - S3 = 11
Mamy a4 = S4 -S3 = 11 oraz r = a5 - a4 = 15 - 11 = 4
r = 4
a4 = a1 +3r
a1 =a4 - 3r = 11 - 3*4 = 11 - 12 = -1
a1 = -1
an =a1 +(n-1)*r = -1 + (n- 1) *4 = -1 - 4 + 4n = 4n - 5
an = 4n - 5

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2009-11-13T20:31:12+01:00
A)a3=14 i S=30
Z sumy: 30=((a1+14):2)*3
30=(3a1+42):2
3a1+42=60
a1=6 wzór an=6+(n-1)*4=4n+2
c) a5=15
S4-S3=11
z różnicy wynika że a4=11
r=15-11=4
kolejne wyrazy ciagu to: a1=-1, a2=3, a3=7, a4=11, a5=15
wzór: an=-1 + (n-1) *4=4n-5
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