Question

S3=40 si S6=60 progresie geometrica bn>1. Sa se afle S9

Answer 1

Vezi atas, te rog!
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Succes!

Answer 2

$\it S_6 -S_3 = 60-40 \Rightarrow b_1+b_2+ b_3+ b_4+b_5+b_6 - (b_1+b_2+b_3)= \\\;\\=20 \Rightarrow b_4+b_5+b_6 =20 \Rightarrow b_4+b_4q+b_4q^2=20 \Rightarrow \\\;\\ \Rightarrow b_4(1+q+q^2) = 20\ \ \ \ (1) \\\;\\ S_3=40 \Rightarrow b_1+b_2+b_3 =40 \Rightarrow b_1+b_1q+b_1q^2=40 \Rightarrow \\\;\\ \Rightarrow b_1(1+q+q^2)=40\ \ \ \ (2)$

$\it (1), (2) \Rightarrow \dfrac{b_4}{b_1}=\dfrac{1}{2} \Rightarrow b_4=\dfrac{b_1}{2} \ \ \ \ (3)$

$\it (3) \Rightarrow b_7=b_4q^3=\dfrac{b_1q^3}{2} = \dfrac{b_4}{2} \\\;\\ b_8=b_7q=\dfrac{b_4q}{2} =\dfrac{b_5}{2} \\\;\\ b_9=b_8q=\dfrac{b_5q}{2} =\dfrac{b_6}{2}$

$\it b_7+b_8+b_9= \dfrac{b_4+b_5+b_6}{2} = \dfrac{20}{2} = 10 \\\;\\ S_9=S_6+b_7+b_8+b_9 = 60 + 10 = 70$