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MCbelko
@MCbelko
July 2022
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явл ли число 21 членом арифмет прогрессии, в которой А1=4,2;А15=5,4
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sangers
A₁=4,2 a₁₅=5,4
a₁₅=a₁+14d=5,4
4,2+14d=5,4
14d=1,2
d=1,2/14=(6/5):14=3/35
a₁+(n-1)*d=21
4,2+(n-1)*(3/35)=21
21/5+3*n/35-3/35=21
(21*7+3*n-3)/35=21
147+3*n-3=21*35
3*n+144=735
3*n=591 |÷3
n=197 ⇒
Ответ: a₁₉₇=21.
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Answers & Comments
a₁₅=a₁+14d=5,4
4,2+14d=5,4
14d=1,2
d=1,2/14=(6/5):14=3/35
a₁+(n-1)*d=21
4,2+(n-1)*(3/35)=21
21/5+3*n/35-3/35=21
(21*7+3*n-3)/35=21
147+3*n-3=21*35
3*n+144=735
3*n=591 |÷3
n=197 ⇒
Ответ: a₁₉₇=21.