Answer:
37. [tex]\frac{3^{144} }{4^{141} }[/tex]
38. 151,7 cm
Step-by-step explanation:
QUESTION 38
We have a geometric progression, with ration equals (3/4)³ :
[tex]a_n=a_1((\frac{3}{4} )^{3} )^{n-1} \\\\a_n=a_1(\frac{3}{4} )^{3n-3}[/tex]
The problem says that: [tex]a_2=128[/tex]
Using the equation to find a1 (ANSWER OF QUESTION 38)
[tex]128=4^{3}*2 =a_1(\frac{3}{4} )^{3*2-3} \\\\4^{3}*2=a_1(\frac{3}{4} )^{3} \\\\a_1=\frac{4^{6} }{27} \\\\a_1=151,7 centimeters\\\\[/tex]
QUESTION 37
find the a_50
[tex]a_5_0=(\frac{4^{6} }{27}) .(\frac{3}{4} )^{(3*49)} \\\\a_5_0=\frac{3^{144} }{4^{141} }[/tex]
a_50 approximately ZERO
