{an}是公差为-2的等差数列,a1+a4+a7+....+a28=90,求a4+a6+a8+.....+a22?
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a1+a4+a7+....+a28=90
(a1+a28)*10/2=90
a1+a28=18
a28=a1+27d=a1-54
a1=36.
a4=a1+3d=36-6=30
a22=a1+21d=36-42=-6
a4+a6+a8+.....+a22
=(a4+a22)*10/2
=120
a1+a4+a7+....+a28
=(a1+a28)*10/2=90
所以
a1+a28=18
a4+a22=a1+a25=a1+a28+6=18+6=24
a4+a6+a8+.....+a22
=(a4+a22)*10/2
=24*10/2
=120
式子1:
a1+a4+a7+....+a28=(a1)+(a1+3d)+(a1+6d)+...+(a1+27d)
=10a1+(d+3d+6d+...27d)
=10a1+135d
=90
式子2:
a4+a6+a8+.....+a22=(a1+3d)+(a1+5d)+(a1+7d)+...+(a1+21d)
=10a1+(3d+5d+7d+...21d)
=10a1+120d
所以式子1=90-15d=90-15*(-2)=120
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