Question

1. Să se determine primii doi termeni ai progresiei geometrice (bn) dacă :
a) b8=256, q=4;
b) b10= - 4 supra 27 (minusul este în dreptul liniei de fracție ) , q= - 1 supra 3 (tot la fel, minusul este în dreptul liniei de fracție )

2. Fie (bn) o progresie geometrică. Să se determine suma Sn a primilor n termeni dacă :
a) progresie geometrică 9, -3, 1,.. și n=7
b) b1=2,5 q=1,5 n=5
c) b11 supra b2=512, n=10, S3=21 supra 2
d) b5=123b2, n=6, S4= 312

Repede , vă rog.

Answer 1

$\displaystyle 1a).b_8=256,~q=4 \\ \\ b_8=256 \Rightarrow b_1 \cdot 4^{8-1}=256 \Rightarrow b_1 \cdot 4^7=256 \Rightarrow b_1 \cdot 4^7=4^4 \Rightarrow \\ \\ \Rightarrow b_1=4^4:4^7 \Rightarrow b_1=4^{4-7} \Rightarrow b_1=4^{-3} \Rightarrow b_1= \frac{1}{4^3} \Rightarrow \boxed{b_1= \frac{1}{64}} \\ \\ b_2=b_1 \cdot 4^{2-1} \Rightarrow b_2=b_1 \cdot 4^1 \Rightarrow b_2= \frac{1}{64} \cdot 4 \Rightarrow b_2= \frac{4}{64} \Rightarrow \boxed{b_2= \frac{1}{16} }$
$\displaystyle b). b_{10}=- \frac{4}{27} ,~q=- \frac{1}{3} \\ \\ b_{10}=- \frac{4}{27} \Rightarrow b_1 \cdot \left(- \frac{1}{3} \right)^{10-1}=- \frac{4}{27} \Rightarrow b_1 \cdot \left(- \frac{1}{3} \right)^9=- \frac{4}{27} \Rightarrow \\ \\ \Rightarrow b_1= \frac{- \frac{4}{27} }{\left(- \frac{1}{3} \right)^9} \Rightarrow b_1=- \frac{4}{27} \cdot (-3)^9 \Rightarrow b_1=- \frac{4}{27} \cdot (-19683) \Rightarrow \\ \\ \Rightarrow b_1= \frac{78732}{27}\Rightarrow\boxed{b_1=2916}$
$\displaystyle b_2=2916 \cdot \left(- \frac{1}{3} \right)^{2-1} \Rightarrow b_2=2916 \cdot \left (- \frac{1}{3} \right)^1 \Rightarrow b_2=2916 \cdot \left(- \frac{1}{3} \right) \Rightarrow \\ \\ \Rightarrow b_2=- \frac{2916}{3} \Rightarrow \boxed{b_2=-972}$
$\displaystyle 2a).9,~-3,~1,...~~~~n=7 \\ \\ b_1=9 ~~~~~~~~~~~~~~~~~~~q=- \frac{3}{9} \Rightarrow q=- \frac{1}{3} \\ \\ S_7=9 \cdot \frac{\left(- \frac{1}{3} \right)^7-1}{- \frac{1}{3} -1} \Rightarrow S_7=9 \cdot \frac{- \frac{1^7}{3^7} -1}{- \frac{1}{3} - \frac{3}{3} } \Rightarrow S_7=9 \cdot \frac{- \frac{1}{2187} -1}{- \frac{4}{3} } \Rightarrow$
$\displaystyle \Rightarrow S_7=9 \cdot \frac{- \frac{1}{2187} - \frac{2187}{2187} }{- \frac{4}{3} } \Rightarrow S_7=9 \cdot \frac{- \frac{2188}{2187} }{- \frac{4}{3} } \Rightarrow \\ \\ \Rightarrow S_7=9 \cdot \left(- \frac{2188}{2187} \right):\left(- \frac{4}{3} \right) \Rightarrow S_7=9 \cdot \left(- \frac{2188}{2187} \right) \cdot \left(- \frac{3}{4} \right) \Rightarrow \\ \\ \Rightarrow S_7=9 \cdot \frac{6564}{8748} \Rightarrow S_7= \frac{6564}{972} \Rightarrow \boxed{S_7= \frac{547}{81} }$
$\displaystyle b).b_1=2,5,~q=1,5,~n=5 \\ \\ b_1=2,5 \Rightarrow b_1= \frac{25}{10} \Rightarrow b_1= \frac{5}{2} \\ \\ q=1,5 \Rightarrow q= \frac{15}{10} \Rightarrow q= \frac{3}{2} \\ \\ S_5= \frac{5}{2} \cdot \frac{\left( \frac{3}{2} \right)^5-1}{ \frac{3}{2} -1} \Rightarrow S_5= \frac{5}{2} \cdot \frac{ \frac{3^5}{2^5}-1 }{ \frac{3}{2} - \frac{2}{2} } \Rightarrow S_5= \frac{5}{2} \cdot \frac{ \frac{243}{32}-1 }{ \frac{1}{2} } \Rightarrow$
$\displaystyle \Rightarrow S_5= \frac{5}{2} \cdot \frac{ \frac{243}{32}- \frac{32}{32} }{ \frac{1}{2} } \Rightarrow S_5= \frac{5}{2} \cdot \frac{ \frac{211}{32} }{ \frac{1}{2} } \Rightarrow S_5= \frac{5}{2} \cdot \frac{211}{32} : \frac{1}{2} \Rightarrow \\ \\ \Rightarrow S_5= \frac{5}{2} \cdot \frac{211}{32} \cdot 2 \Rightarrow S_5= \frac{5}{2} \cdot \frac{211}{16} \Rightarrow \boxed{S_5= \frac{1055}{32} }$
$\displaystyle c). \frac{b_{11}}{b_2} =512,~n=10,~S_3= \frac{21}{2} \\ \\ \frac{b_{11}}{b_2} =512 \Rightarrow \frac{b_1 \cdot q^{11-1}}{b_1 \cdot q^{2-1}} =512 \Rightarrow \frac{b_1 \cdot q^{10}}{b_1 \cdot q} =512 \Rightarrow q^9=512 \Rightarrow \\ \\ \Rightarrow q^9=2^9 \Rightarrow q=2$
$\displaystyle S_3= \frac{21}{2} \Rightarrow b_1 \cdot \frac{2^{3}-1}{2-1} = \frac{21}{2} \Rightarrow b_1 \cdot (8-1)= \frac{21}{2} \Rightarrow \\ \\ \Rightarrow b_1 \cdot 7= \frac{21}{2} \Rightarrow b_1= \frac{ \frac{21}{2} }{7} \Rightarrow b_1= \frac{21}{2} :7 \Rightarrow b_1= \frac{21}{2} \cdot \frac{1}{7} \Rightarrow \\ \\ \Rightarrow b_1= \frac{21}{14} \Rightarrow b_1= \frac{3}{2}$
$\displaystyle S_{10}= \frac{3}{2} \cdot \frac{2^{10}-1}{2-1} \Rightarrow S_{10}= \frac{3}{2} \cdot (1024-1) \Rightarrow S_{10}= \frac{3}{2} \cdot 1023\Rightarrow \\ \\ \Rightarrow \boxed{S_{10}= \frac{3069}{2} }$