Se dau numerele pozitive a1, a2, a3 in progresie aritmetica. Daca a1+a2+a3=21 si numerele a1+2, a2+3 si a3+9 sunt in progresie geometrica, atunci suma a1^2+a2^2+a3^2 =?
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a2 +3 = (a1 +2)q (a1 + r + 3) = (a1 + 2)q
a3 + 9 = (a2 + 3)q (a1 + 2r+9) = (a1 + r +3)q
(a1 +r + 3)/(a1 + 2r + 9) = (a1 + 2)/(a1 + r +3)
(a1 + r + 3)² = (a1 + 2r + 9)(a1 + 2)
a1² + r² + 9 + 2a1r + 6a1 + 6r = a1² + 2a1r + 11a1 + 4r + 18
r² + 2r - 5a1 = 9
r² + 2r-5(7 - r) - 9 = 0 r² +7r - 44 = 0 r² -4r + 11r - 44 = 0
r(r - 4) + 11(r - 4) = 0 (r-4)(r+11) = 0 r - 4 = 0 r = 4 r = 11 = 0 r = -11
a1 = 3 si a1 = 18
3² +7² + 11² = 179
18² + 7² + (-4)² = 324 + 49 + 16 = 389