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This is a Korean CSAT mathematics multiple-choice problem. Solve it and put only the correct choice symbol, one of ①, ②, ③, ④, ⑤, inside \boxed{}.
Problem 12 Let{a n}be a geometric sequence satisfying 2(a1 +a 4 +a 7) =a 4 +a 7 +a 10 = 6. Determine the value ofa 10. [4 points] ① 22 7 ② 24 7 ③ 26 7 ④ 30 7 ⑤ 32 7
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로그인 회원가입 You are solving a Korean CSAT mathematics problem. Reason carefully, but the final response must contain exactly one final answer inside \boxed{}.
For multiple-choice problems, put only one of ①, ②, ③, ④, ⑤ inside \boxed{}. For short-answer problems, put only the final integer inside \boxed{}. Do not use external tools.
Problem 12 Let{a n}be a geometric sequence satisfying 2(a1 +a 4 +a 7) =a 4 +a 7 +a 10 = 6. Determine the value ofa 10. [4 points] ① 22 7 ② 24 7 ③ 26 7 ④ 30 7 ⑤ 32 7 더보기
Let the common ratio be 𝑟 r. Then
𝑎 4
𝑎 1 𝑟 3 , 𝑎 7
𝑎 1 𝑟 6 , 𝑎 10
𝑎 1 𝑟 9 . a 4
=a 1
r 3 ,a 7
=a 1
r 6 ,a 10
=a 1
r 9 .
From 2 ( 𝑎 1 + 𝑎 4 + 𝑎 7 )
6 2(a 1
+a 4
+a 7
)=6 we have
𝑎 1 + 𝑎 4 + 𝑎 7
a 1
+a 4
+a 7
=3.
From 𝑎 4 + 𝑎 7 + 𝑎 10
6 a 4
+a 7
+a 10
=6,
𝑎 4 + 𝑎 7 + 𝑎 10
𝑟 3 ( 𝑎 1 + 𝑎 4 + 𝑎 7 )
𝑟 3 ⋅ 3
6 , a 4
+a 7
+a 10
=r 3 (a 1
+a 4
+a 7
)=r 3 ⋅3=6,
so 𝑟 3
2 r 3 =2. Then
𝑎 1 + 𝑎 4 + 𝑎 7
𝑎 1 ( 1 + 𝑟 3 + 𝑟 6 )
𝑎 1 ( 1 + 2 + 4 )
7 𝑎 1
3 , a 1
+a 4
+a 7
=a 1
(1+r 3 +r 6 )=a 1
(1+2+4)=7a 1
=3,
giving 𝑎 1
3 7 a 1
= 7 3
. Therefore
𝑎 10
𝑎 1 𝑟 9
𝑎 1 ( 𝑟 3 ) 3
3 7 ⋅ 8
24 7 . a 10
=a 1
r 9 =a 1
(r 3 ) 3
7 3
⋅8= 7 24
.
This matches option ②.
② ②
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Comparative Analysis (P11) (Average across runs)
Averages use reported measurements only. Missing usage is not treated as zero.