Question

1. Determinati primul termen al progresiei geometrice b1, 6, b3, 24.

2. Determinati primul termen si ratia unei progresii aritmetice stiind ca a1+a2=3 si a3+a4=12.

Answer 1

1. b3²=6*24=144⇒b3=12
b3=b2*q⇒q=b3/b2=12/6=2
b1=b2/q=6/2=3
2. a1+a2=a1+a1+r= 2a1+r=3
 a3+a4=a1+2r+a1+3r= 2a1+5r=12
$\left \{ {{2a1+r=3} \atop {2a1+5r=12}} \right.$
Daca le scazi obtii 4r=9⇒ r=9/4
2a1+r=3
a1=(3-r)/2=$\frac{3- \frac{9}{4} }{2} = \frac{ \frac{12-9}{4} }{2}= \frac{3}{8}$