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dkfdshjdgss
@dkfdshjdgss
August 2022
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В арифметической прогрессии сумма первых четырех членов прогрессии равна 12, а сумма первых восьми членов равна 40. Найдите сумму первых одиннадцати членов этой прогрессии
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Маша888
Sn = 1/2(2a1+(n-1)*d)*n
S4 = 1/2 (2a1+3d)*4 = 4a1+6d = 12
S8 = 1/2 (2a1+7d)*8 = 8a1+28d = 40
Система:
4a1+6d=12
8a1+28d=40
16d = 16
d=1
4a1+6=12
4a1=6
a1 = 3/2 = 1.5
S11-?
S11= 1/2 (2a1+10d)*11 = 1/2 (3+10)*11 = 13*11/2 = 71.5
Ответ: 71.5
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Answers & Comments
S4 = 1/2 (2a1+3d)*4 = 4a1+6d = 12
S8 = 1/2 (2a1+7d)*8 = 8a1+28d = 40
Система:
4a1+6d=12
8a1+28d=40
16d = 16
d=1
4a1+6=12
4a1=6
a1 = 3/2 = 1.5
S11-?
S11= 1/2 (2a1+10d)*11 = 1/2 (3+10)*11 = 13*11/2 = 71.5
Ответ: 71.5