Question

Sa se determine primii doi termeni ai progresiei geometrice (bn) daca:
a) b8=256 , q=4
b) b9=768 , b11=3072
c) b10= -4/27 , q= -1/3
Dau coroana

Answer 1

[tex]\displaystyle a).b_8=256,~q=4 \\ b_8=256 \Rightarrow b_1 \cdot 4^{8-1}=256 \Rightarrow b_1 \cdot 4^7=256 \Rightarrow b_1 \cdot 4^7=4^4 \Rightarrow \\ \Rightarrow b_1=4^4:4^7 \Rightarrow b_1=4^{4-7} \Rightarrow b_1=4^{-3} \Rightarrow b_1= \frac{1}{4^3} \Rightarrow \boxed{b_1= \frac{1}{64}} \\ b_2=b_1 \cdot 4^{2-1} \Rightarrow b_2=b_1 \cdot 4^1 \Rightarrow b_2= \frac{1}{64} \cdot 4 \Rightarrow b_2= \frac{4}{64} \Rightarrow \boxed{b_2= \frac{1}{16} } [/tex]
$\displaystyle b). \left \{ {{b_9=768} \atop {b_{11}=3072}} \right. \Rightarrow \left \{ {{b_1 \cdot q^{9-1}=768} \atop {b_1 \cdot q^{11-1}=3072}} \right. \Rightarrow \left \{ {{b_1 \cdot q^8=768}\atop{b_1 \cdot q^{10}=3072}}\right.\Rightarrow\\ \Rightarrow\frac{b_1 \cdot q^1^0}{b_1 \cdot q^{8}} = \frac{3072}{768} \Rightarrow q^2=4 \Rightarrow q= \sqrt{4} \Rightarrow q=2,~q=-2\\b_1\cdot q^8=768\Rightarrow b_1\cdot2^8=768\Rightarrow b_1 \cdot 256=768\Rightarrow b_1=\frac{768}{256}\Rightarrow \boxed{b_1=3}$
$\displaystyle b_2=3 \cdot 2^{2-1} \Rightarrow b_2=3 \cdot 2^1 \Rightarrow b_2=3 \cdot 2 \Rightarrow \boxed{b_2=6} \\ b_2=3 \cdot (-2)^{2-1} \Rightarrow b_2=3 \cdot (-2)^1 \Rightarrow b_2=3 \cdot (-2) \Rightarrow \boxed{b_2=-6}$
$\displaystyle c). b_{10}=- \frac{4}{27} ,~q=- \frac{1}{3} \\ b_{10}=- \frac{4}{27} \Rightarrow b_1 \cdot \left(- \frac{1}{3} \right)^{10-1}=- \frac{4}{27} \Rightarrow b_1 \cdot \left(- \frac{1}{3} \right)^9=- \frac{4}{27} \Rightarrow \\ \Rightarrow b_1= \frac{- \frac{4}{27} }{\left(- \frac{1}{3} \right)^9} \Rightarrow b_1=- \frac{4}{27} \cdot (-3)^9 \Rightarrow b_1=- \frac{4}{27} \cdot (-19683) \Rightarrow \\ \Rightarrow b_1= \frac{78732}{27} \Rightarrow \boxed{b_1=2916}$
$\displaystyle b_2=2916 \cdot \left(- \frac{1}{3} \right)^{2-1} \Rightarrow b_2=2916 \cdot \left (- \frac{1}{3} \right)^1 \Rightarrow b_2=2916 \cdot \left(- \frac{1}{3} \right) \Rightarrow \\ \Rightarrow b_2=- \frac{2916}{3} \Rightarrow \boxed{b_2=-972}$