please help !!! In an arithmetic sequence, t(8) = 145 and t(50) = 607. Find t(129).

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

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please help !!! In an arithmetic sequence, t(8) = 145 and t(50) = 607. Find t(129).

$$\small{\text{$ \begin{array}{lrcl} & t_8 = a+(8-1)d &=& a+7d \\ & t_{50} =a + (50-1)d &=& a+49 d\\ \\ (1) & \dfrac{ t_{50}+t_8 }{2} &=& a + 28d \\ \\ & t_{50}-t_8 &=& 42d \\ (2) & ( t_{50}-t_8 )\cdot \dfrac{100}{42} &=& 100d\\ \\ & t_{129} &=& a + (129-1)d \\ & t_{129} &=& a + 128d \\ & t_{129} &=& a + 28d + 100d \qquad | \qquad a + 28d = \dfrac{ t_{50}+t_8 }{2} \qquad 100d =( t_{50}-t_8 )\cdot \dfrac{100}{42} \\ & t_{129} &=& \dfrac{ t_{50}+t_8 }{2} + ( t_{50}-t_8 )\cdot \dfrac{100}{42}\\\\ & t_{129} &=& \dfrac{607+145}{2} + (607-145)\dfrac{100}{42}\\\\ & t_{129} &=& 376 + 1100 \\ &\mathbf{ t_{129} }& \mathbf{=} & \mathbf{ 1476 }\\ \end{array} $}}$$