text: +0
  
  # please help !!! In an arithmetic sequence, t(8) = 145 and t(50) = 607. Find t(129).
  
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  $$\underset{\,\,\,\,{{\rightarrow {\mathtt{in, an, arithmetic, sequence, and, find}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{In}}={\mathtt{0}}\\ {\mathtt{an}}={\mathtt{0}}\\ {\mathtt{arithmetic}}={\mathtt{0}}\\ {\mathtt{sequence}}={\mathtt{0}}\\ {t}{\left({\mathtt{8}}\right)}={\mathtt{0}}\\ \mathrm{ERROR}={\mathtt{145}}\\ {\mathtt{and}}={\mathtt{0}}\\ {t}{\left({\mathtt{50}}\right)}={\mathtt{0}}\\ \mathrm{ERROR}={\mathtt{607}}\\ {\mathtt{Find}}={\mathtt{0}}\\ \mathrm{ERROR}={\mathtt{0}}\end{array}\right)}$$
  
  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