W ciągu arytmetycznym a2+a5=19 i a3=8. Wyznacz a1 i r.
Z góry dzięki.
a₃=8
a₂+a₅=19
a₂=a₃-r
a₅=a₃+2r
a₃-r+a₃+2r=19
2a₃+r=19
2·8+r=19
16+r=19
r=3
a₁=a₃-2r
a₁=8-2·3
a₁=8-6
a₁=2
a2+a5=19 a1+r+a1+4r=19
a3=8 a1+2r=8
a1=8-2r
8-2r+r+8-2r+4r=19
a1=2
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a₃=8
a₂+a₅=19
a₂=a₃-r
a₅=a₃+2r
a₃-r+a₃+2r=19
2a₃+r=19
2·8+r=19
16+r=19
r=3
a₁=a₃-2r
a₁=8-2·3
a₁=8-6
a₁=2
a2+a5=19 a1+r+a1+4r=19
a3=8 a1+2r=8
a1=8-2r
8-2r+r+8-2r+4r=19
r=3
a1=2