Question

Salut, am nevoie de ajutor la ex. 2.540A

Answer 1

Răspuns:

$\text{f)} a_7 = 1458$

Explicație pas cu pas:

$a_1 + a_2 + a_3 = 26\\ \\ a_1 + a_1\cdot q + a_1\cdot q^2 = 26\\ \\ a_1(q^2 + q + 1) = 26\implies a_1= \frac{26}{q^2 + q + 1}\\ \\ a_5 + a_6 + a_7 = 2106\\ \\ a_1\cdot q^4 + a_1 \cdot q^5 + a_1\cdot q^6 = 2106\\ \\ a_1\cdot q^4(1 + q + q^2) = 2106\\ \\ \frac{26}{q^2 + q + 1}\cdot q^4(q^2+q+1) = 2106\\ \\ 26q^4 = 2106\\ \\ q^4 = 81\Bigg | \sqrt[4]{()} \\ \\ q = \pm\sqrt[4]{81}\\ \\ q = \pm 3\\ \\ \text{Stim ca termenii sunt pozitivi, deci q trebuie sa fie pozitiv} \\ \\ q = 3\implies a_1 = \frac{26}{3^2+3+1} = \frac{26}{9+4} = \frac{26}{13} = 2\\ \\ a_7 = a_1\cdot q^6 = 2\cdot 3^6 = 2\cdot 9^3 = 2\cdot 81\cdot 9 = 2\cdot 729 = 1458$