text: +0
  
  # please help !!! In an arithmetic sequence, t(8) = 145 and t(50) = 607. Find t(129).
  
  0
  559
  2
  
  $$\underset{\,\,\,\,{{\rightarrow {\mathtt{in, an, arithmetic, sequence, and, find}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{In}}={\mathtt{0}}\\ {\mathtt{an}}={\mathtt{0}}\\ {\mathtt{arithmetic}}={\mathtt{0}}\\ {\mathtt{sequence}}={\mathtt{0}}\\ {t}{\left({\mathtt{8}}\right)}={\mathtt{0}}\\ \mathrm{ERROR}={\mathtt{145}}\\ {\mathtt{and}}={\mathtt{0}}\\ {t}{\left({\mathtt{50}}\right)}={\mathtt{0}}\\ \mathrm{ERROR}={\mathtt{607}}\\ {\mathtt{Find}}={\mathtt{0}}\\ \mathrm{ERROR}={\mathtt{0}}\end{array}\right)}$$
  
  Jul 25, 2015
  
  #1
  +27342
  +10
  
  Arithmetic sequence is t(n) = a + (n-1)*d  where a is the first term, d is the difference and t(n) is the n'th term
  
  So:
  
  145 = a + 7d  from the 8'th term
  
  607 = a + 49d from the 50'th term
  
  Subtract the first equation from the second
  
  462 = 42d
  
  d = 462/42 = 11
  
  Substitute this back into the first equation
  
  145 = a + 7*11
  
  a = 145 - 77 = 68
  
  Hence t(129) = 68 + 128*11 = 1476
  
  .
  
  Jul 25, 2015
  
  #1
  +27342
  +10
  
  Arithmetic sequence is t(n) = a + (n-1)*d  where a is the first term, d is the difference and t(n) is the n'th term
  
  So:
  
  145 = a + 7d  from the 8'th term
  
  607 = a + 49d from the 50'th term
  
  Subtract the first equation from the second
  
  462 = 42d
  
  d = 462/42 = 11
  
  Substitute this back into the first equation
  
  145 = a + 7*11
  
  a = 145 - 77 = 68
  
  Hence t(129) = 68 + 128*11 = 1476
  
  .
  
  Alan Jul 25, 2015
  #2
  +20831
  +5
  
  $$\small{\text{ \begin{array}{lrcl} & t_8 = a+(8-1)d &=& a+7d \\ & t_{50} =a + (50-1)d &=& a+49 d\\ \\ (1) & \dfrac{ t_{50}+t_8 }{2} &=& a + 28d \\ \\ & t_{50}-t_8 &=& 42d \\ (2) & ( t_{50}-t_8 )\cdot \dfrac{100}{42} &=& 100d\\ \\ & t_{129} &=& a + (129-1)d \\ & t_{129} &=& a + 128d \\ & t_{129} &=& a + 28d + 100d \qquad | \qquad a + 28d = \dfrac{ t_{50}+t_8 }{2} \qquad 100d =( t_{50}-t_8 )\cdot \dfrac{100}{42} \\ & t_{129} &=& \dfrac{ t_{50}+t_8 }{2} + ( t_{50}-t_8 )\cdot \dfrac{100}{42}\\\\ & t_{129} &=& \dfrac{607+145}{2} + (607-145)\dfrac{100}{42}\\\\ & t_{129} &=& 376 + 1100 \\ &\mathbf{ t_{129} }& \mathbf{=} & \mathbf{ 1476 }\\ \end{array} }}$$
  
  Jul 27, 2015
explanation: The document contains a math question and an answer to this question. The question is about finding the 129th term of an arithmetic sequence given the 8th and 50th terms. The answer is provided by two users, both of whom solve the problem and arrive at the same result, \( t(129) = 1476 \).
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: +0
  
  # please help !!! In an arithmetic sequence, t(8) = 145 and t(50) = 607. Find t(129).
  
  0
  559
  2
  
  $$\underset{\,\,\,\,{{\rightarrow {\mathtt{in, an, arithmetic, sequence, and, find}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{In}}={\mathtt{0}}\\ {\mathtt{an}}={\mathtt{0}}\\ {\mathtt{arithmetic}}={\mathtt{0}}\\ {\mathtt{sequence}}={\mathtt{0}}\\ {t}{\left({\mathtt{8}}\right)}={\mathtt{0}}\\ \mathrm{ERROR}={\mathtt{145}}\\ {\mathtt{and}}={\mathtt{0}}\\ {t}{\left({\mathtt{50}}\right)}={\mathtt{0}}\\ \mathrm{ERROR}={\mathtt{607}}\\ {\mathtt{Find}}={\mathtt{0}}\\ \mathrm{ERROR}={\mathtt{0}}\end{array}\right)}$$
  
  Jul 25, 2015
  
  #1
  +27342
  +10
  
  Arithmetic sequence is t(n) = a + (n-1)*d  where a is the first term, d is the difference and t(n) is the n'th term
  
  So:
  
  145 = a + 7d  from the 8'th term
  
  607 = a + 49d from the 50'th term
  
  Subtract the first equation from the second
  
  462 = 42d
  
  d = 462/42 = 11
  
  Substitute this back into the first equation
  
  145 = a + 7*11
  
  a = 145 - 77 = 68
  
  Hence t(129) = 68 + 128*11 = 1476
  
  .
  
  Jul 25, 2015
  
  #1
  +27342
  +10
  
  Arithmetic sequence is t(n) = a + (n-1)*d  where a is the first term, d is the difference and t(n) is the n'th term
  
  So:
  
  145 = a + 7d  from the 8'th term
  
  607 = a + 49d from the 50'th term
  
  Subtract the first equation from the second
  
  462 = 42d
  
  d = 462/42 = 11
  
  Substitute this back into the first equation
  
  145 = a + 7*11
  
  a = 145 - 77 = 68
  
  Hence t(129) = 68 + 128*11 = 1476
  
  .
  
  Alan Jul 25, 2015
  #2
  +20831
  +5
  
  $$\small{\text{ \begin{array}{lrcl} & t_8 = a+(8-1)d &=& a+7d \\ & t_{50} =a + (50-1)d &=& a+49 d\\ \\ (1) & \dfrac{ t_{50}+t_8 }{2} &=& a + 28d \\ \\ & t_{50}-t_8 &=& 42d \\ (2) & ( t_{50}-t_8 )\cdot \dfrac{100}{42} &=& 100d\\ \\ & t_{129} &=& a + (129-1)d \\ & t_{129} &=& a + 128d \\ & t_{129} &=& a + 28d + 100d \qquad | \qquad a + 28d = \dfrac{ t_{50}+t_8 }{2} \qquad 100d =( t_{50}-t_8 )\cdot \dfrac{100}{42} \\ & t_{129} &=& \dfrac{ t_{50}+t_8 }{2} + ( t_{50}-t_8 )\cdot \dfrac{100}{42}\\\\ & t_{129} &=& \dfrac{607+145}{2} + (607-145)\dfrac{100}{42}\\\\ & t_{129} &=& 376 + 1100 \\ &\mathbf{ t_{129} }& \mathbf{=} & \mathbf{ 1476 }\\ \end{array} }}$$
  
  Jul 27, 2015
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and an answer to this question. The question is about finding the 129th term of an arithmetic sequence given the 8th and 50th terms. The answer is provided by two users, both of whom solve the problem and arrive at the same result, \( t(129) = 1476 \).
  
  Label: question-and-answer