В арифметической прогрессии (аn), а5 = 10, а11 = 40. Найдите а8?. Created by Mariaa77. algebra-ru

В арифметической прогрессии (аn), а5 = 10, а11 = 40. Найдите а8?...

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$a_5=10; a_{11}=40$
$a_n=a_1+(n-1)*d$
$a_5=a_1+(5-1)*d=a_1+4d$
$a_{11}=a_1+(11-1)*d=a_1+10d$
$a_{11}-a_5=(a_1+10d)-(a_1+4d)=a_1+10d-a_1-4d=6d$
$a_{11}-a_5=40-10=30$
$6d=30$
$d=30:6=5$
$a_1=a_5-4d$
$a_1=10-4*5=-10$
$a_8=-10+(8-1)*5=25$
ответ: 25

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Pacanka202 A11=a1+10d
а5=а1+4d
a1+10d=40
a1+4d=10
6d=30
d=5
a1=10–4•5=–10
a8=a1+7d=–10+7•5=25

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