Question

Se considera o progresie geometrica bn, cu ratia q=2. Determinati n natural pentru care bn=96, iar suma primilor n termeni ai progresiei este 189.

Answer 1

[tex]\it S_n=\dfrac{b_nq-b_1}{q-1} \Rightarrow 189=\dfrac{96\cdot2-b_1}{2-1} \Rightarrow 189= 192-b_1 \Rightarrow \\\;\\ \\\;\\ \Rightarrow b_1=192-189 \Rightarrow b_1 = 3[/tex]

[tex]\it b_n=b_1\cdot q^{n-1} \Rightarrow 96 = 3\cdot2^{n-1}|_{:3} \Rightarrow 32=2^{n-1} \Rightarrow 2^5=2^{n-1}\Rightarrow \\\;\\ \Rightarrow n-1=5 \Rightarrow n=6[/tex]