4. Найдите сумму первых пяти членов геометрической прогрессии (bₙ), в которой: a) b₄ = 1/16, b₅ = 1/64; б) b₂ = 4, b₄ = 36, q > 0.

a) \(q = \frac{b_5}{b_4} = \frac{\frac{1}{64}}{\frac{1}{16}} = \frac{1}{64} \cdot 16 = \frac{1}{4}\)

\(b_4 = b_1 \cdot q^3 \Rightarrow b_1 = \frac{b_4}{q^3} = \frac{\frac{1}{16}}{(\frac{1}{4})^3} = \frac{\frac{1}{16}}{\frac{1}{64}} = \frac{1}{16} \cdot 64 = 4\)

\[S_5 = \frac{b_1(1-q^5)}{1-q} = \frac{4(1-(\frac{1}{4})^5)}{1-\frac{1}{4}} = \frac{4(1-\frac{1}{1024})}{\frac{3}{4}} = \frac{4 \cdot \frac{1023}{1024}}{\frac{3}{4}} = \frac{\frac{1023}{256}}{\frac{3}{4}} = \frac{1023}{256} \cdot \frac{4}{3} = \frac{341}{64} = 5.328125\]

б) \(b_4 = b_2 \cdot q^2 \Rightarrow q^2 = \frac{b_4}{b_2} = \frac{36}{4} = 9 \Rightarrow q = \pm 3\). Так как \(q > 0\), то \(q = 3\)

\(b_2 = b_1 \cdot q \Rightarrow b_1 = \frac{b_2}{q} = \frac{4}{3}\)

\[S_5 = \frac{b_1(1-q^5)}{1-q} = \frac{\frac{4}{3}(1-3^5)}{1-3} = \frac{\frac{4}{3}(1-243)}{-2} = \frac{\frac{4}{3} \cdot (-242)}{-2} = \frac{-\frac{484}{3}}{-2} = \frac{484}{3} \cdot \frac{1}{2} = \frac{242}{3} = 80\frac{2}{3} \approx 80.67\]

Ответ: а) 341/64 = 5.328125; б) 242/3 ≈ 80.67