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Auzhan
@Auzhan
August 2022
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{an} арифметическая прогрессия
а1=4,2
а10=15,9
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novo51
An=a1+(n-1)d
d=(an-a1)/(n-1)=(a10-a1)/(10-1)=(15,9-4,2)/9=11,7/9=1,3
Sn=(2a1+(n-1)d)/2*n
S15=(2*4,2+(15-1)*1,3)/2*15=(8,4+14*1,3)/2*15=(4,2+7*1,3)*15=(4,2+9,1)*15=
=13,3*15=199,5
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Answers & Comments
d=(an-a1)/(n-1)=(a10-a1)/(10-1)=(15,9-4,2)/9=11,7/9=1,3
Sn=(2a1+(n-1)d)/2*n
S15=(2*4,2+(15-1)*1,3)/2*15=(8,4+14*1,3)/2*15=(4,2+7*1,3)*15=(4,2+9,1)*15=
=13,3*15=199,5