Wyznacz ciąg arytmetyczny (an)...- Zadanie 11: MATeMAtyka 3. Zakres rozszerzony. Po gimnazjum

Rozwiązanie

$a)$

${a_{1} + a_{2} = - 6 a_{2} + a_{4} = 0$

${a_{1} + a_{1} + r = - 6 a_{1} + r + a_{1} + 3 r = 0$

${2 a_{1} + r = - 6 ∣ \cdot (- 1) 2 a_{1} + 4 r = 0$

${- 2 a_{1} - r = 6 2 a_{1} + 4 r = 0$

$- r + 4 r = 6 + 0$

$3 r = 6$

$r = 2$

$- 2 a_{1} - r = 6$

$- 2 a_{1} - 2 = 6$

$- 2 a_{1} = 8$

$a_{1} = - 4$

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

$a_{n} = - 4 + (n - 1) \cdot 2$

$a_{n} = - 4 + 2 n - 2$

$a_{n} = 2 n - 6$

$b)$

${a_{1} + a_{3} = 2 a_{2} \cdot a_{4} = 2$

${a_{1} + a_{1} + 2 r = 2 (a_{1} + r) \cdot (a_{1} + 3 r) = 2$

${2 a_{1} + 2 r = 2 ∣ : 2 a_{1}^{2} + 3 a_{1} r + a_{1} r + 3 r^{2} = 2$

${a_{1} + r = 1 ∣ - r a_{1}^{2} + 4 a_{1} r + 3 r^{2} = 2$

${a_{1} = 1 - r (1 - r)^{2} + 4 (1 - r) r + 3 r^{2} = 2$

$(1 - r)^{2} + 4 (1 - r) r + 3 r^{2} = 2$

$1 - 2 r + r^{2} + 4 r - 4 r^{2} + 3 r^{2} = 2$

$2 r = 1$

$r = \frac{1}{2}$

$a_{1} = 1 - \frac{1}{2}$

$a_{1} = \frac{1}{2}$

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

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

$a_{n} = \frac{1}{2} + \frac{1}{2} n - \frac{1}{2}$

$a_{n} = \frac{1}{2} n$

$c)$

${a_{3} \cdot a_{5} = - 8 a_{3} + a_{7} = - 2$

${(a_{1} + 2 r) \cdot (a_{1} + 4 r) = - 8 a_{1} + 2 r + a_{1} + 6 r = - 2$

${a_{1}^{2} + 4 a_{1} r + 2 a_{1} r + 8 r^{2} = - 8 2 a_{1} + 8 r = - 2 ∣ : 2$

${a_{1}^{2} + 6 a_{1} r + 8 r^{2} = - 8 a_{1} + 4 r = - 1 ∣ - 4 r$

${a_{1}^{2} + 6 a_{1} r + 8 r^{2} = - 8 a_{1} = - 1 - 4 r$

${(- 1 - 4 r)^{2} + 6 (- 1 - 4 r) r + 8 r^{2} = - 8 a_{1} = - 1 - 4 r$

$(- 1 - 4 r)^{2} + 6 (- 1 - 4 r) r + 8 r^{2} = - 8$

$1 + 8 r + 16 r^{2} - 6 r - 24 r^{2} + 8 r^{2} = - 8$

$2 r = - 9$

$r = - 4, 5$

$a_{1} = - 1 - 4 r$

$a_{1} = - 1 - 4 \cdot (- 4, 5)$

$a_{1} = - 1 + 18$

$a_{1} = 17$

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

$a_{n} = 17 + (n - 1) \cdot (- 4, 5)$

$a_{n} = 17 - 4, 5 n + 4, 5$

$a_{n} = 21, 5 - 4, 5 n$

Magdalena Matusik

Nauczycielka matematyki

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