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07 Jun 2015, 09:01
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\(t_1, \ t_2, \ t_3, \ ..., \ t_n, \ ... \)
In the sequence above, each term after the first term is obtained by adding a constant \(k\) to the preceding term. If \(t_1+ t_3+t_5+t_7=32\), what is the value of \(t_2+t_4+t_6\)?
A. 8
B. 12
C. 24
D. 32
E. 72
In the sequence above, each term after the first term is obtained by adding a constant \(k\) to the preceding term. If \(t_1+ t_3+t_5+t_7=32\), what is the value of \(t_2+t_4+t_6\)?
A. 8
B. 12
C. 24
D. 32
E. 72
07 Jun 2015, 09:01
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Official Solution:
\(t_1, \ t_2, \ t_3, \ ..., \ t_n, \ ... \)
In the sequence above, each term after the first term is obtained by adding a constant \(k\) to the preceding term. If \(t_1+ t_3+t_5+t_7=32\), what is the value of \(t_2+t_4+t_6\)?
A. 8
B. 12
C. 24
D. 32
E. 72
According to the stem:
\(t_2=t_1+k\);
\(t_3=t_2+k=t_1+2k\);
\(t_4=t_3+k=t_1+3k\);
...
\(t_n=t_1+(n-1)k\);
Given that \(t_1 + t_3 + t_5 + t_7 = 32\), we have \(t_1 + (t_1 + 2k) + (t_1 + 4k) + (t_1 + 6k) = 32\). Thus, \(t_1 + 3k = 8\).
We need to find the value of \(t_2 + t_4 + t_6 = (t_1 + k) + (t_1 + 3k) + (t_1 + 5k) = 3(t_1 + 3k)\). From the previous result, we know that \(t_1 + 3k = 8\), so \(3(t_1 + 3k) = 3 * 8 = 24\).
Answer: C
\(t_1, \ t_2, \ t_3, \ ..., \ t_n, \ ... \)
In the sequence above, each term after the first term is obtained by adding a constant \(k\) to the preceding term. If \(t_1+ t_3+t_5+t_7=32\), what is the value of \(t_2+t_4+t_6\)?
A. 8
B. 12
C. 24
D. 32
E. 72
According to the stem:
\(t_2=t_1+k\);
\(t_3=t_2+k=t_1+2k\);
\(t_4=t_3+k=t_1+3k\);
...
\(t_n=t_1+(n-1)k\);
Given that \(t_1 + t_3 + t_5 + t_7 = 32\), we have \(t_1 + (t_1 + 2k) + (t_1 + 4k) + (t_1 + 6k) = 32\). Thus, \(t_1 + 3k = 8\).
We need to find the value of \(t_2 + t_4 + t_6 = (t_1 + k) + (t_1 + 3k) + (t_1 + 5k) = 3(t_1 + 3k)\). From the previous result, we know that \(t_1 + 3k = 8\), so \(3(t_1 + 3k) = 3 * 8 = 24\).
Answer: C
21 Jul 2016, 06:08
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I think this is a high-quality question and I agree with explanation. Nice manipulative question
