Question

Wyznacz 1 wyraz i iloraz oraz podaj wzór ogółny c.geometrycznego,wiedząc że: a)a1+a3=10

a2+a4=5

b)a3=-4

a4=1/4

c)a2+a3=3/4

a2-a4=3/8

Answer 1

WYZNACZ WYRAZ POCZĄTKOWY I ILORAZ ORAZ PODAJ OGÓLNY WZÓR CIĄGU GEOMETRYCZNEGO, WIEDZĄC, ŻE:
a) a2=1 i a5=1/8
Korzystam ze wzoru na n-ty wyraz ciagu geometrycznego
an = a1*q(do potegi n-1)
a2 = a1*q = 1
a5 = a1*q⁴ = 1/8

a1*q = 1
a1*q⁴ = 1/8

a1 = 1: q
(1:q)*q⁴ = 1/8

a1 = 1 :q
q³ = 1/8

a1 = 1:q
q = ∛(1/8)


a1 = 1 :q
q = 1/2

a1 = 1 : 1/2 = 1*2 =2
q =1/2

a1= 2
q = 1/2

Obliczam wyraz ogólny ciagu geometrycznego an
an = a1*q(do potegi n-1)
an = 2* (1/2)(do potegi n-1)
an = 2*(1/2) do ptęgi n : 1/2
an = 2*(1/2) do ptęgi n *2
an = 4*(1/2) do potegi n

b)a3=-4 i a4=1/4
a3 = a1*q² = -4
a4 = a1*q₃ = 1/4

q = a4: a3
q = (1/4):(-4)
q = (1/4)*(-4)
q = -1

a1*q² = -4
a1* (-1)² = -4
a1 *1 = -4
a1 = (-4) :1
a1 = -4

Obliczam wyraz ogólny ciagu geometrycznego an
an = a1*q(do potegi n-1)
an = (-4)*(-1)do potegi(n-1)
an = (-4)*(-1) do potegi n : (-1)
an = 4*(-1) do potegi n

c)a1 +a3=10 i a2+a4=5

a1 + a1*q² = 10
a1*q + a1*q³ = 5

a1(1 + q²) = 10
a1*q (1 + q²) = 5

a1 = 10 : (1 + q²)
[10:(1 +q²)]*q *(1+q²) = 5

a1 = 10 : (1 + q²)
10*q = 5

a1 = 10 : ( 1+q²)
q = 5 /10 = 1/2

a1 = 10 : [ 1 + (1/2)²] = 10 : [ 1 + 1/4] = 10 : 5/4 = 10*4/5 = 8
q = 1/2

a1 = 8
q = 1/2


aObliczam wyraz ogólny ciagu geometrycznego an
n = a1*q(do potegi n-1)
an = 8*(1/2)(do potegi n-1)
an = 8*(1/2)*do potęgi n : 1/2
an = 16*(1/2) so potęgi n

d)a2+a4=10 i a3-a2=2

a1*q² - a1*q = 2
a1*q + a1*q³ = 10

a1*q (q -1)= 2
a1*q (1 +q²) = 10

a1*q = 2 : (q -1)
a1*q = 10 : (1 + q²)

porównuję stronami
2 : (q -1) = 10 : (1 + q²)
2(1 + q²) = 10*(q -1)
2 + 2q² = 10q -10
2q² -10q +2 +10 = 0
2q² -10q +12 = 0 /:2
q² -5q +6 = 0
Δ = (-5)² - 4*1*6 = 25 -24 = 1
√Δ = √1 =1
q1= (5 -1): 2*1 = 4 :2 = 2
q2 = (5 +1) : 2*1 = 6 : 2 = 3

Obliczam teraz a1 dla q = 2
a1*q = 2 : (q -1)
a1 *2 = 2 : (2 -1 )
a1*2 = 2 : 1
a1*2 = 2 /:2
a1 = 2 :2
a1 = 1

Teraz obliczam a1 dla q = 3
a1 q = 2 : (q -1)
a1*3 = 2 : (3 -1)
a1*3 = 2 : 2
a1*3 = 1
a1 = 1/3

a1 = 1 i q = 2, lub a1 = 1/3 i q = 3

Obliczam wyraz ogólny ciagu geometrycznego an
an= a1*q(do potegi n-1)
an = 1*2 do potegi n-1
an = 1*2 do potegi n :2
an = 1/2*2 do potegi n

lub
an= a1*q(do potegi n-1)
an = 1/3*3 do potegi n-1
an= 1/3*3 do potegi n : 3
an = 1/9*3 do potegi n

e)a2+a3=3/4 i a2-a4=3/8

a1*q + a1*q² = 3/4
a1*q - a1*q³ = 3/8

a1*q(1 + q) = 3/4
a1*q( 1 -q²) = 3/8

a1*q = 3/4: ( 1 +q)
a1*q = 3/8 : ( 1 -q²)

porównuję stronami
3/4 : ( 1 +q) = 3/8: ( 1- q²)
3/4(1 - q²) = 3/8*(1 + q)
3/4 - 3/4q² = 3/8 + 3/8q
-3/4q² -3/8q = 3/8 -3/4
-3/4q² - 3/8q = 3/8 - 6/8
-3/4q² - 3/8q + 3/8 = 0 /*(-8)
6q² + 3q - 3 = 0 /:3
2q² +q -1 = 0

Δ = 1² - 4*2*(-1) = 1 + 8 = 9
√Δ =√ 9 = 3

q1 = (-1 -3): 2*2 = (-4): 4 = -1
q2 = (-1 +3) : 2*2 = 2 : 4 = 1/2

Obliczam a1 dla q = -1

a1*q = 3/4: ( 1 +q)
a1*(-1) = 3/4 : (1 -1)
a1*(-1) = 3/4 : 0 - brak rozwiazania

dla q = -1 brak rozwiazania
Obliczam a1 dla q = 1/2
a1*q = 3/4 : (1 +q)
a1*1/2 = 3/4 : ( 1 + 1/2)
a1*1/2 = 3/4 : (3/2)
a1*1/2 = 3/4 *(2/3)
a1*1/2 = 1/2
a1 = 1/2 : 1/2 =1
a1 = 1

a1 = 1
q = 1/2

Obliczam wyraz ogólny ciagu geometrycznego an
an= a1*q(do potegi n-1)
an = 1*(1/2) do potęgi n -1
an = 1*(1/2) do potęgi n : (1/2)
an = 2*(1/2) do potęgi n