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# Penn writes a 2013-term arithmetic sequence of positive integers, and Teller writes a different 2013-term arithmetic sequence of integers. T

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Penn writes a 2013-term arithmetic sequence of positive integers, and Teller writes a different 2013-term arithmetic sequence of integers. Teller's first term is the negative of Penn's first term. Each then finds the sum of the terms in his sequence. If their sums are equal, then what is the smallest possible value of the first term in Penn's sequence?

Guest Jan 17, 2015

### Best Answer

#4
+19653
+10

Penn writes a 2013-term arithmetic sequence of positive integers, and Teller writes a different 2013-term arithmetic sequence of integers. Teller's first term is the negative of Penn's first term. Each then finds the sum of the terms in his sequence. If their sums are equal, then what is the smallest possible value of the first term in Penn's sequence?

Arithmetic series Penn: $$\small{ \text{ p_n = p_1 + (n-1) d_p  }}\ . \quad \text{The sum} \ \text{  s_p= \frac{n}{2}*[2p_1+(n-1)d_p]  }}$$

Arithmetic series Teller: $$\small{\text{t_n = t_1 + (n-1) d_t  }}\ . \quad \text{The sum} \ \text{  s_t= \frac{n}{2}*[2t_1+(n-1)d_t]  }}$$

I.

$$\small{\text{  s_p=s_t  }} \\ \small{\text{ \frac{n}{2}*[2p_1+(n-1)d_p]=\frac{n}{2}*[2t_1+(n-1)d_t]  }} \\ \small{\text{ 2p_1+(n-1)d_p = 2t_1+(n-1)d_t  }}$$

II.

$$\small{\text{  t_1= -p_1  }} \\ \small{\text{ 2p_1+(n-1)d_p = -2p_1 +(n-1)d_t  }} \\ \small{\text{ 4p_1 = (n-1)d_t - (n-1)d_p = (n-1)(d_t-d_p)  }}$$

III.

$$\small{\text{  n=2013 }} \\ \small{\text{ 4p_1 = 2012(d_t-d_p) }} \\ \small{\text{ p_1 = 503(d_t-d_p) }}$$

IV.

The smallest possible value of the first term in  Penn's sequence $$\small{\text{p_1}}$$ is 503, if $$\small{\text{(d_t-d_p) = 1 }}$$

heureka  Jan 19, 2015
#1
+808
0

I'm not sure I get it. Both Penn and Teller have sequences with positive integers, yet Teller's first term is the negative of Penn's first term. That sounds contradicting to me ...?

Tetration  Jan 17, 2015
#2
+92805
-1

No Tetration.  Tellers sequence is integers but they do not have to be positive.

you are right of course - they cannot all be positive.

Melody  Jan 18, 2015
#3
+92805
-1

I have drawn more attention to this question by posting a reminder here

http://web2.0calc.com/questions/this-unanswered-question-has-been-bugging-me

Melody  Jan 19, 2015
#4
+19653
+10
Best Answer

Penn writes a 2013-term arithmetic sequence of positive integers, and Teller writes a different 2013-term arithmetic sequence of integers. Teller's first term is the negative of Penn's first term. Each then finds the sum of the terms in his sequence. If their sums are equal, then what is the smallest possible value of the first term in Penn's sequence?

Arithmetic series Penn: $$\small{ \text{ p_n = p_1 + (n-1) d_p  }}\ . \quad \text{The sum} \ \text{  s_p= \frac{n}{2}*[2p_1+(n-1)d_p]  }}$$

Arithmetic series Teller: $$\small{\text{t_n = t_1 + (n-1) d_t  }}\ . \quad \text{The sum} \ \text{  s_t= \frac{n}{2}*[2t_1+(n-1)d_t]  }}$$

I.

$$\small{\text{  s_p=s_t  }} \\ \small{\text{ \frac{n}{2}*[2p_1+(n-1)d_p]=\frac{n}{2}*[2t_1+(n-1)d_t]  }} \\ \small{\text{ 2p_1+(n-1)d_p = 2t_1+(n-1)d_t  }}$$

II.

$$\small{\text{  t_1= -p_1  }} \\ \small{\text{ 2p_1+(n-1)d_p = -2p_1 +(n-1)d_t  }} \\ \small{\text{ 4p_1 = (n-1)d_t - (n-1)d_p = (n-1)(d_t-d_p)  }}$$

III.

$$\small{\text{  n=2013 }} \\ \small{\text{ 4p_1 = 2012(d_t-d_p) }} \\ \small{\text{ p_1 = 503(d_t-d_p) }}$$

IV.

The smallest possible value of the first term in  Penn's sequence $$\small{\text{p_1}}$$ is 503, if $$\small{\text{(d_t-d_p) = 1 }}$$

heureka  Jan 19, 2015
#5
+87309
+3

Very nice, heureka...........

CPhill  Jan 19, 2015
#6
+92805
+2

Thanks very much Heureka

Melody  Jan 19, 2015
#7
+92805
+7

I missed the bit about there being 2013 terms    doh!

Melody  Jan 19, 2015

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