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determinati suma primilor 6 termeni ai progresiei geometrice daca b2=6 si b5=48

Răspuns :

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Vezi imaginea LUCSEMBOURG23
b2=b1q
b5=b1q^4
b5/b1=48/6=8=q³⇒q=∛8=2
⇒b1=b2/q=6/2=3
atunci b1+b2+...+b6=3(1+2+..2^5)=3*(2^6-1)/(2-1)=3*63=189
as simple as that!