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greenwalley
@greenwalley
July 2022
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Найдите сумму тридцати трёх первых членов арифметической прогрессии (an),если а3+а5+а13=33 и а15-а8-а10=-1.
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mikael2
Verified answer
A3+a5+a13=3a1+2d+4d+12d=3a1+18d
3a1+18d=33
a1+6d=11 (1)
a15-a8-a10=-a1+14d-7d-9d=-a1-2d
-a1-2d=-1
a1+2d=1 (2)
вычтем из 1 равенство 2
6d-2d=11-1 4d=10 d=10/4=2.5
a1=1-2d=1-5=-4
s33=[2*(-4)+2.5*32]*33/2=72*33/2=1188
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Answers & Comments
Verified answer
A3+a5+a13=3a1+2d+4d+12d=3a1+18d3a1+18d=33 a1+6d=11 (1)
a15-a8-a10=-a1+14d-7d-9d=-a1-2d
-a1-2d=-1 a1+2d=1 (2)
вычтем из 1 равенство 2
6d-2d=11-1 4d=10 d=10/4=2.5
a1=1-2d=1-5=-4
s33=[2*(-4)+2.5*32]*33/2=72*33/2=1188