Question

W pewnym ciągu geometrycznym suma pierwszego i trzeciego wyrazu jest równa 25 a suma drugiego i czwartego 30.podaj piąty wyraz

Answer 1

a1+a3=25

a2+a4=30

 

a1+a1*q²=25

a1*q+a1*q³=30

 

a1(1+q²)=25

a1(q+q³)=30

 

a1=25/(1+q²)

a1=30/(q+q³)

 

25/(1+q²)=30/(q+q³)

25q³-30q²+25q-30=0

5q³-6q²+5q-6=0

q²(5q-6)+(5q-6)=0

(5q-6)(q²+1)=0

5q-6=0

5q=6

q=6/5

 

a1= 25/(1+q²)=25/(1+(6/5)²)=25/(1+36/25)=25/(25/25+36/25)=25/(61/25)=

=25*25/61=625/61

 

a5=a1*q⁴=625/61*(6/5)⁴=625/61*1296/625=1296/61

 

licze na naj 

Answer 2

a1 + a3 = 25

a2 + a4 = 30

czyli

a1 + a1*q^2 = 25

a1*q + a1*q^3 = 30

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a1*(1 + q^2) = 25  => a1 = 25 / ( 1 + q^2)

[ 25 / ( 1 + q^2)]*q + [ 25/ (1 + q^2)] *q^3 = 30  / * ( 1 + q^2)

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25 q + 25 q^3 = 30 + 30* q^2

25 q^3 - 30 q^2 + 25 q - 30 = 0  / : 5

5 q^3 - 6 q^2 + 5 q - 6 = 0

5 q*( q^2 + 1) - 6*(q^2 + 1) = 0

( 5 q - 6)*(q^2 + 1) = 0

5 q - 6 = 0 ;  q^2 + 1 > 0

5 q = 6

q = 6/5

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a1 = 25 /[ 1 + (6/5)^2] = 25 / [ 25/25 + 36/25] = 25 / (61/25) =625/61

zatem

a5 = a1 *q^4 = [ 625/61) * (6/5)^4 = (625/61)*(1296/625) = 1296/61

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