Question

Wyznacz sumę n poczatkowych wyrazow ciagu arytmetycznego (an) o roznicy r i wyrazie pierwszym a₁.
b)a₁=-10 r=3 n=20
c)a₁=50 r=-½ n=17
d)a₁=4,35 r=0 n=100


oblicz sume kolejnych wyrazow ciagu arytmetycznego.
b)2+6+10...+202

Pilne dam najlepsze za wszystko zrobione

Answer 1

B)a₁=-10 r=3 n=20
a₂₀=a₁+(n-1)r
a₂₀=47
Sn=((a₁+an)/2)*n
S₂₀=869,5

c)a₁=50 r=-½ n=17
a₁₇=42
S₁₇=((a₁+a₁₇)/2)17=782

d)a₁=4,35 r=0 n=100
S₁₀₀=400
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2+6+10...+202
a₁=2 an=202 r=4
an=a₁+(n-1)r
202=2+4n-4
204=4n
n=51
a₅₁=202
Sn=((a₁+an)/2)n
S₅₁=5202

Answer 2

B)a₁=-10 r=3 n=20
an = a1 + (n-1)*r
a20 = -10 + ( 20 - 1)*3
a20 = -10 + 19*3
a20 = -10 + 57
a20 = 47

Sn = 1/2(a₁ + an)*n
S20= 1/2(-10 +47)*20
S20= 1/2*37*20
S20 = 370

c)a₁=50 r=-½ n=17
an = a1 + (n-1)*r
a17 = 50 + (17-1)*(-1/2)
a17= 50 + 16*(-1/2)
a17= 50 -8
a17 = 42

Sn = 1/2(a₁ + an)*n
S17 = 1/2(50 +42)*17
S17 = 1/2* 92*17
S17 = 782

d)a₁=4,35 r=0 n=100
an = a1 + (n-1)*r
a100 = 4,35 + (100-1)*0
a100= 4,35 + 99*0
a100= 4,35 + 0
a100= 4,35

Sn = 1/2(a₁ + an)*n
S100 = 1/2(4,35 +4,35)*100
S100 = 1/2(8,7)*100
S100= 435


oblicz sume kolejnych wyrazow ciagu arytmetycznego.
b)2+6+10...+202
a1 = 2
a2= 6
a3 = 10
r = a2-a1 = const.
r = 6-2 = 4
an = 202
an= a1 + (n-1)*r
an = 2 + (n-1)*4 = 202
2 + 4n -4 = 202
4n -2 = 202
4n = 202 +2
4n = 204
n = 204 : 4
n = 51
Sn = 1/2(a₁ + an)*n
S51 = 1/2(2 + 202)*51
S51 = 1/2*204*51
S51 = 5202