Question

Zad 1 Oblicz pierwszy wyraz ciągu geometrycznego , w którym a3= 10 , a5= 1/3.

Zad 2 Suma kolejnych potęg liczby 3 od 3 do potegi 1 do 3 do potegi 9 jest równa...? proszę o pomoc:(

Answer 1

1.

q²=1/3:10=1/30

a3/a1=q²

a1=a3/q²

a1=10:1/30=300

2.

a1=3

q=3

n=9

Sn=a1(1-q^n)/(1-q)

 

Sn=3(1-3^9)/(1-3)=3*(1-19683)/(-2)=3*9841= 29523

Answer 2

z.1

a3 = 10

a5 = 1/3

Korzystamy z wzoru:

an = a1*q^(n-1)

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Mamy

a3 = a1*q^2 = 10

a5 = a1*q^4 = 1/3

Dzielimy stronami

a5 /a3 = [ a1 *q^4 ]/[ a1*q^2] = ( 1/3) : 10

q^2 = 1/30

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a1*( 1/30) = 10

a1 = 10 / (1/30) = 10* 30 = 300

Odp. a1 = 300

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z.2

a1 = 3^1 = 3

a2 = 3^2 = 9

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a9 = 3^9

a2 : a1 = 9 : 3 = 3

zatem q = 3

Sn = a1*[ 1 - q^n]/[ 1 - q ]

czyli

S9 = 3*[1 - 3^9]/ [ 1 - 3] = 3*[ 1 - 19 683 ]/( - 2) = -1,5 * ( - 19 682 ) = 29 523

Odp. S9 = 29 523

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