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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?

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 $$\small{\text{$(d_t-d_p) = 1 $}}$$

@@ End of Day Wrap   Mon 19/1/15     Sydney, Australia      Time 6:05pm

Hi everyone,

It has been 2 days since I did the last wrap and since then there has been some fantastic answers from Tetration, Rosala, CPhill, geno3141, kingpifireplasma, Tenacious, Mathematician, Sasini, alan and Nauseated.  Thank you all.  

Interest posts:

1) This is why people must learn to use brackets properly.   Thanks Tetration and Melody

Tues 20/1/15

1) This tricky AP question had Chris and I stumped.  Heureka came to our rescue   Thanks Heureka.  

$$\\If cos \theta = 0.34\\
then\\
\theta = inverse\; cos\; of\; 0.34\\
$Inverse cos of 0.34 is often written as $cos^{-1}(0.34)\\
$Arc cos is exactly the same as inverse cos so on the web2.0calc this can be entered as $
[2nd][acos](0.34)=$$

I am just typing into the input box so i dont need the [2nd] I will enter     acos(0.34)=

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\mathtt{0.34}}\right)} = {\mathtt{70.123\: \!125\: \!929\: \!921^{\circ}}}$$

Original sequence   a=a1    d=d

$$S_n=\frac{n}{2}[2a+(n-1)d]\qquad $This is the sum of an AP$$$

Solve simultaneously

(2)-(1)

d=3-17/5 = -2/5

sub into (2)

a₁+5(-2/5)=3

a₁-2=3

a₁=5

need to find the one equivalent to n+3n+2n(squared) using distributive property

2n(n+2)

2n(n+1)

n(n+2)

2n(2n+2)

$$n+3n+2n^2 \\
= n(1+3+2n) \\
= n ( 4+2n ) \\
= n * 2(2+n) \\
=2n(n+2)$$

Do you mean "how much less?"

There really is not enough info given here.   The question does not make much sense to me.

$$P(Exactly\; 2\;hits)=5C2*0.95^2*0.05^3$$

$${\left({\frac{{\mathtt{5}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{5}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{0.95}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{{\mathtt{0.05}}}^{{\mathtt{3}}} = {\mathtt{0.001\: \!128\: \!125}}$$

the opposite of log base 10 is 10^(whatever)

so the antilog of 6 is  10^6=1000000

$$\\4=log_{10}10000\\
so\\
10^4=10000\\$$

differentiate 3x4(x+2)

$$3*4(x+2) \\
=12(x+2) \\
=12x+24 \\ \\
\frac{\ d (12x+24)} { dx } = \frac{\ d (12x)} { dx }+\frac{\ d (24)} { dx } \\ \\
\frac{\ d (12x+24)} { dx } = 12+0 \\ \\
\frac{\ d (12x+24)} { dx } = 12$$

Consider the function g(x)=3. Find g(2).

It does not matter what x is g(x) will always equal 3

so

g(2)=3

I have no idea what you are asking.   

Think about this

what is   $$17-3^2 \;\;\;?$$

Well I expect that you would agree it is 17-9 = 8

so..... $$-3^2=-9$$      How can this be justified?

WELL

$$(-3)^2=-3*-3=+9$$

BUT

$$-3^2=-1*3^2=-1*3*3=-9$$

So there is an invisable   1*    between the minus sign and the 3.

That is how it can be justified!

So to summerize

$$\\-3^2=-9\\
(-3)^2=+9$$

0+0+9/1*10/5*100/5*10

there is a calc on the home page :)

All fixed, thanks anon :)

(1.9 × 1010) + (3.7 × 109)

I suspect that this is what you are really after     ^  means 'to the power of'

(1.9 × 10^10) + (3.7 × 10^9)

You can put it straight into the site calc OR

=(1.9 × 10 ×10^9) + (3.7 × 10^9)

=(19 ×10^9) + (3.7 × 10^9)

= (19+3.7)×10^9

=22.7×10^9

=2.27×10×10^9

=2.27× 10^10

Oops! Error on line 8 there, Melody.

I know you included that deliberately, just to see who is following along.

log x^3 + 2 log x - 10 = 0

3log x + 2log x = 10

5logx = 10   divide both sides by 5

log x = 2     ..so...in exponential terms...

10^2  = x  = 100

Here's another way to do this one, rosala

(l+m)^2 - (l-m)^2

[ (l + m) - (l - m) ] [ (l + m) + (l - m)] =

[2m] [2l ]=

4ml