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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?

Jan 17, 2015

#4
+20850
+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 $$\small{\text{(d_t-d_p) = 1 }}$$

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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 ...?

Jan 17, 2015
#2
+95361
-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.

Jan 18, 2015
#3
+95361
-1

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

Jan 19, 2015
#4
+20850
+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
#5
+94611
+3

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

Jan 19, 2015
#6
+95361
+2

Thanks very much Heureka

Jan 19, 2015
#7
+95361
+7

I missed the bit about there being 2013 terms    doh!

Jan 19, 2015