Odpowiedzi

2010-03-15T14:48:15+01:00
An = a1 + (n-1)r

S5 = 10
S5 = 0,5(a1 + a5)5
S5 = 2,5(a1 + a1 + 4r)
S5 = 2,5(2a1 + 4r)
S5 = 5a1 + 10r
5a1 + 10r = 10 |:5
a1 + 2r = 2
a1 = 2 - 2r

(a3, a5, a13) - ciąg geometryczny
(a5)² = a3 * a13
(a1 + 4r)² = (a1 + 2r)(a1 + 12r)
(a1)² + 8a1 r + 16r² = (a1)² + 12a1 r + 2a1 r + 24r²
6a1 r + 8r² = 0
3a1 r + 4r² = 0

3(2 - 2r)r + 4r² = 0
6r - 6r² + 4r² = 0
-2r² + 6r = 0
r² - 3r = 0
r(r - 3) = 0
r = 0 lub r - 3 = 0
r = 0 lub r = 3

jeśli r = 0, to a1 = 2 - 2r = 2
an = 2 + (n-1)0
an = 2 ----> wzór ogólny ciągu, w którym a1 = 2 i r = 0

jeśli r = 3, to a1 = 2 - 2r = 2 - 6 = -4
an = -4 + (n-1)3
an = -4 + 3n - 3
an = 3n - 7 ---> wzór ogólny ciągu, w którym a1 = -4 i r = 3

Zadanie to posiada dwa rozwiązania.
3 3 3