+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
The text passage contains a mathematical question about an arithmetic sequence and provides a solution to the problem.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
+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
user
The text passage contains a mathematical question about an arithmetic sequence and provides a solution to the problem.

TYPE: SINGLE_QUESTION_WITH_ANSWER