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Dianocika
@Dianocika
September 2021
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10 термен арифметической прогрессиии равен 19,а сумма первых 50 терменов прогрессии равна 2500. Найти сумму 3,12,20 термена.
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ndusha
Verified answer
А10=а1+9д
19=а1+9д
9д=19-а1
д=(19-а1)/9
S50=(2а1+49д)/2 *50=(2а1+49д) *25
2500=
(2а1+49д) *25
2а1+49д=2500/25
2а1+49д=100
49д=100-2а1
д=(100-2а1)/49
(19-а1)/9=
(100-2а1)/49
49(19-а1)=9
(100-2а1)
931-49а1=900-18а1
-49а1+18а1=900-931
-31а1=-31
а1=-31/(-31)
а1=1
д=(19-а1)/9
д=(19-1)/9=18/9=2
а3=1+2*2=5
а12=1+11*2=23
а20=1+19*2=39
а3+а12+а20
=5+23+39=
67
2 votes
Thanks 1
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Answers & Comments
Verified answer
А10=а1+9д19=а1+9д
9д=19-а1
д=(19-а1)/9
S50=(2а1+49д)/2 *50=(2а1+49д) *25
2500=(2а1+49д) *25
2а1+49д=2500/25
2а1+49д=100
49д=100-2а1
д=(100-2а1)/49
(19-а1)/9=(100-2а1)/49
49(19-а1)=9(100-2а1)
931-49а1=900-18а1
-49а1+18а1=900-931
-31а1=-31
а1=-31/(-31)
а1=1
д=(19-а1)/9
д=(19-1)/9=18/9=2
а3=1+2*2=5
а12=1+11*2=23
а20=1+19*2=39
а3+а12+а20=5+23+39=67