Question

Va rog daca puteti sa rezolvati oricare multumesc.​

Answer 1

Răspuns:

progresie aritmetică, progresie geometrică

Explicație pas cu pas:

$a_{n} = a_{1} + (n - 1) \cdot r$

$S_{n} = \frac{(a_{1} + a_{n})\cdot n}{2}$

19.

$S_{12} = \frac{(a_{1} + a_{12})\cdot 12}{2} = \frac{(1 + 111)\cdot 12}{2} = 112\cdot6 = 672$

20.

$2(2x - 3) = (x + 1) + (x - 3)$

$4x - 6 = 2x - 2$

$2x = 4 = > x = 2$

21.

$r = a_{2} - a_{1} = 5 - 2 = 3$

$a_{6} = 2 + (6 - 1) \cdot 3 = 2 + 15 = 17$

$S_{6} = \frac{(a_{1} + a_{6})\cdot 6}{2} = \frac{(2 + 17)\cdot 6}{2} = 19\cdot 3 = 57$

22.

$x + 7 = \sqrt{(5 - x)(3x + 11)}$

condiții de existență:

$(5 - x)(3x + 11) \geqslant 0$

$= > x \in \left[ - \frac{11} {3} ; 5\right]$

→

$(x + 7)^{2} = (5 - x)(3x + 11)$

${x}^{2} + 14x + 49 = 15x + 55 - 3 {x}^{2} - 11x$

$4 {x}^{2} + 10x - 6 = 0 \\ 2 {x}^{2} + 5x - 3 = 0$

$(2x - 1)(x + 3) = 0$

$2x - 1 = 0 = > x = \frac{1}{2}$

$x + 3 = 0 = > x = - 3$

=>

$S = \{ - 3 ; \frac{1}{2} \}$

23.

$S_{2} = \frac{(a_{1} + a_{2})\cdot 2}{2}$

$S_{2} = a_{1} + a_{2} = a_{1} + a_{1} + r = 2a_{1} + r$

$10 = 2a_{1} + 4 = > a_{1} = 3$

24.

$\frac{b_{1}}{b_{4}} = \frac{1}{8} <=> b_{4} = b_{1}\cdot q^{3}$

$\frac{b_{1}}{b_{1}\cdot q^{3}} = \frac{1}{2^{3}} <=> \frac{1}{q^{3}} = \frac{1}{2^{3}} => q = 2$

$b_{1} = \frac{b_{2}}{q} => b_{1} = \frac{3}{2}$