Răspuns:
progresie aritmetică, progresie geometrică
Explicație pas cu pas:
[tex]a_{n} = a_{1} + (n - 1) \cdot r[/tex]
[tex]S_{n} = \frac{(a_{1} + a_{n})\cdot n}{2} \\ [/tex]
19.
[tex]S_{12} = \frac{(a_{1} + a_{12})\cdot 12}{2} = \frac{(1 + 111)\cdot 12}{2} = 112\cdot6 = 672 \\ [/tex]
20.
[tex]2(2x - 3) = (x + 1) + (x - 3)[/tex]
[tex]4x - 6 = 2x - 2[/tex]
[tex]2x = 4 = > x = 2[/tex]
21.
[tex]r = a_{2} - a_{1} = 5 - 2 = 3[/tex]
[tex]a_{6} = 2 + (6 - 1) \cdot 3 = 2 + 15 = 17[/tex]
[tex]S_{6} = \frac{(a_{1} + a_{6})\cdot 6}{2} = \frac{(2 + 17)\cdot 6}{2} = 19\cdot 3 = 57 \\ [/tex]
22.
[tex]x + 7 = \sqrt{(5 - x)(3x + 11)}[/tex]
condiții de existență:
[tex](5 - x)(3x + 11) \geqslant 0[/tex]
[tex] = > x \in \left[ - \frac{11} {3} ; 5\right] \\ [/tex]
→
[tex](x + 7)^{2} = (5 - x)(3x + 11)[/tex]
[tex]{x}^{2} + 14x + 49 = 15x + 55 - 3 {x}^{2} - 11x[/tex]
[tex]4 {x}^{2} + 10x - 6 = 0 \\ 2 {x}^{2} + 5x - 3 = 0 [/tex]
[tex](2x - 1)(x + 3) = 0[/tex]
[tex]2x - 1 = 0 = > x = \frac{1}{2} \\ [/tex]
[tex]x + 3 = 0 = > x = - 3[/tex]
=>
[tex]S = \{ - 3 ; \frac{1}{2} \} \\ [/tex]
23.
[tex]S_{2} = \frac{(a_{1} + a_{2})\cdot 2}{2} \\ [/tex]
[tex]S_{2} = a_{1} + a_{2} = a_{1} + a_{1} + r = 2a_{1} + r \\ [/tex]
[tex]10 = 2a_{1} + 4 = > a_{1} = 3[/tex]
24.
[tex]\frac{b_{1}}{b_{4}} = \frac{1}{8} <=> b_{4} = b_{1}\cdot q^{3}\\ [/tex]
[tex]\frac{b_{1}}{b_{1}\cdot q^{3}} = \frac{1}{2^{3}} <=> \frac{1}{q^{3}} = \frac{1}{2^{3}} => q = 2\\ [/tex]
[tex]b_{1} = \frac{b_{2}}{q} => b_{1} = \frac{3}{2} \\[/tex]
