1.w 20-wyrazowym ciągu arytmetycznym iloczyn wyrazów skrajnych =123 a suma wyrazu drugiego i czwartego =14.wyznacz ten ciąg.
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
a₁*a₂₀ = 123
a₂+a₄ = 14
a₂₀ = a₁+(20-1)*r = a₁+19r
a₂ = a₁+(2-1)*r = a₁+r
a₄ = a₁+(4-1)*r = a₁+3r
układ równań:
a₁*(a₁+19r) = 123
a₁+r+a₁+3r = 14
-----------------
a₁²+19a₁r = 123
2a₁+4r = 14 |:2
-----------------
a₁²+19a₁r = 123
a₁+2r = 7 - wyznaczam a₁ -> a₁ = 7-2r
-- do I: (7-2r)²+19r*(7-2r) = 123
49-28r+4r²+133r-38r² = 123
105r-34r² = 74
34r²-105r+74 = 0
Δ = (-105)²-4*34*74
Δ = 11025-10064
Δ = 961
√Δ = 31
r₁ = [-(-105)-31]/(2*34)
r₁ = 74:68
r₁ = 37/34
r₂ = [-(-105)+31]/(2*34)
r₂ = 136:68
r₂ = 2
dla r₁:
a₁+74/34 = 7
a₁ = 4i14/17
wzór: a n = 4i14/17+(n-1)*37/34 = 3i25/34 + 1i3/34*n
n = 1, 2, 3, ..., 20
dla r₂:
a₁+2*2 = 7
a₁ = 3
wzór: a n = 3+(n-1)*2 = 3+2n-2 = 2n+1
n = 1, 2, 3, ..., 20
Przekształcam drugie równanie:
a_{2}
Podstawiam do pierwszego:
-
a