Question

1.W ciągu geometrycznym a_{10}= 24, a iloraz wynosi q=\sqrt{2}.

a) znajdz wyraz pierwszy i czawrt tego ciagu.
b) oblicz sume dziesieciu pozostalych poczatkowych wyrazow tego ciagu

Answer 1

a10=24

q=√2

a)

a10=a1*q⁹

24=a1*(√2)⁹

24=a1*16√2

a1=24/(16√2)=6/(4√2)=3/(2√2)=(3√2)/4

 

a4=a1*q³

a4=(3√2)/4*(√2)³=(3√2)/4*2√2=(6*2)/4=12/4=3

 

b)

s10= (a1*(1-q^n))/1-q= [(3√2)/4*(1-(√2)¹⁰)]/(1-√2)

s10= [(3√2)/4*(1-32)]/(1-√2)

s10= [(3√2)/4 *(-31)]/(1-√2)

s10=(-93√2/4)/(1-√2) * (1+√2)/(1+√2)

s10=(-93√2/4-93/2)/1-2 

s10=(-93√2/4-186/4)/-1

s10=93√2/4+186/4

s10=(93√2+186)/4