+0  
 
0
559
2
avatar

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{in, an, arithmetic, sequence, and, find}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{In}}={\mathtt{0}}\\
{\mathtt{an}}={\mathtt{0}}\\
{\mathtt{arithmetic}}={\mathtt{0}}\\
{\mathtt{sequence}}={\mathtt{0}}\\
{t}{\left({\mathtt{8}}\right)}={\mathtt{0}}\\
\mathrm{ERROR}={\mathtt{145}}\\
{\mathtt{and}}={\mathtt{0}}\\
{t}{\left({\mathtt{50}}\right)}={\mathtt{0}}\\
\mathrm{ERROR}={\mathtt{607}}\\
{\mathtt{Find}}={\mathtt{0}}\\
\mathrm{ERROR}={\mathtt{0}}\end{array}\right)}$$

 Jul 25, 2015

Best Answer 

 #1
avatar+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
avatar+27342 
+10
Best Answer

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
avatar+20831 
+5

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

 

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

 

 Jul 27, 2015

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