$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{in, an, arithmetic, sequence, and, find}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{In}}={\mathtt{0}}\\
{\mathtt{an}}={\mathtt{0}}\\
{\mathtt{arithmetic}}={\mathtt{0}}\\
{\mathtt{sequence}}={\mathtt{0}}\\
{t}{\left({\mathtt{8}}\right)}={\mathtt{0}}\\
\mathrm{ERROR}={\mathtt{145}}\\
{\mathtt{and}}={\mathtt{0}}\\
{t}{\left({\mathtt{50}}\right)}={\mathtt{0}}\\
\mathrm{ERROR}={\mathtt{607}}\\
{\mathtt{Find}}={\mathtt{0}}\\
\mathrm{ERROR}={\mathtt{0}}\end{array}\right)}$$
+33614 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
.
+33614 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
.
+26367 please help !!! In an arithmetic sequence, t(8) = 145 and t(50) = 607. Find t(129).
$$\small{\text{$
\begin{array}{lrcl}
& t_8 = a+(8-1)d &=& a+7d \\
& t_{50} =a + (50-1)d &=& a+49 d\\
\\
(1) & \dfrac{ t_{50}+t_8 }{2} &=& a + 28d \\
\\
& t_{50}-t_8 &=& 42d \\
(2) & ( t_{50}-t_8 )\cdot \dfrac{100}{42} &=& 100d\\
\\
& t_{129} &=& a + (129-1)d \\
& t_{129} &=& a + 128d \\
& t_{129} &=& a + 28d + 100d \qquad | \qquad a + 28d = \dfrac{ t_{50}+t_8 }{2} \qquad 100d =( t_{50}-t_8 )\cdot \dfrac{100}{42} \\
& t_{129} &=& \dfrac{ t_{50}+t_8 }{2} + ( t_{50}-t_8 )\cdot \dfrac{100}{42}\\\\
& t_{129} &=& \dfrac{607+145}{2} + (607-145)\dfrac{100}{42}\\\\
& t_{129} &=& 376 + 1100 \\
&\mathbf{ t_{129} }& \mathbf{=} & \mathbf{ 1476 }\\
\end{array}
$}}$$
