A One-Liner on Arithmetic Progression!

Let sequence have $n>1$ consecutive integers starting with $k+1$ $k+1+k+2+...+k+n=nk+\frac{n(n+1)}{2}=100 \\ 2k=\frac{200}{n}-(n+1)$ Now $n$ must be a divisor of $200$ and the expression $\frac{200}{n}-(n+1)$ must be even. It's easy to see that just $n=5, 8, 25, 40, 200$ work.

Let the sum of the consecutive integers be an arithmetic progression with $d = 1$ . Hence, since the sum of those numbers is equal to $100$ then,

$\large{S_n = \frac{n\cdot (a_1 + a_n)}{2} \Rightarrow 100 = \frac{n\cdot \big(2\cdot a_1 + (n - 1)\cdot d\big)}{2}}$ , for $n > 1$ $\Rightarrow n^2 + (2\cdot a_1 - 1)\cdot n -200 = 0$ Let's approach this expression as an parametric quadratic equation. Since we are working in the set of integers then both $a_1$ and $n$ must be integers, which means that the discriminant must be a perfect square of an integer. But, $\large{n =\frac{1 - 2\cdot a_1 \pm \sqrt{Δ}}{2}}$ , so $Δ$ must be a square of an odd number so that $n$ is integer. Thus, $Δ = (2\cdot a_1 - 1)^2 + 800 = (2\cdot r + 1)^2$ , for some integer $r$

$\Rightarrow (2\cdot a_1 - 1)^2 - (2\cdot r + 1)^2 = - 800$

$\Rightarrow (- a_1 + r + 1)\cdot (a_1 + r) = 200 = 2^3\cdot 5^2$

So the possible pairs are:

$\Big((-a_1+r+1), (a_1+r)\Big)=\Big\{(1, 200), (2, 100), (4, 50), (5, 40), (8, 25), (10, 20)\Big\}$

But only the pairs: $\Big\{(1, 200), (5, 40), (8, 25)\Big\}$ give integers $a_1, r$

1)Pair: $(1, 200)$

• $(- a + r + 1) = 1, (a + r) = 200 \Rightarrow a = 100, r = 100$

  So, $n = 1 || n = -200$

• $(- a + r + 1) = 200, (a + r) = 1 \Rightarrow a = -99, r = 100$

  So, $n = 200 || n = -1$

2)Pair: $(5, 40)$

• $(- a + r + 1) = 5, (a + r) = 40 \Rightarrow a = 18, r = 22$

  So, $n = 5 || n = -40$

• $(- a + r + 1) = 40, (a + r) = 5 \Rightarrow a = -17, r = 22$

  So, $n = 40 || n = -5$

3)Pair: $(8, 25)$

• $(- a + r + 1) = 8, (a + r) = 25 \Rightarrow a = 9, r = 16$

  So, $n = 8 || n = -25$

• $(- a + r + 1) = 25, (a + r) = 8 \Rightarrow a = -8, r = 16$

  So, $n = 25 || n = -8$

From the above values of $n$ we disregard all the negative values and the one where $n = 1$ because of the restriction $n > 1$ . If you count them you'll find out $5$ correct values, each one refering to a unique arithmetic progression!

a is the leading term of a possible sequence such that $a<100\in \bf Z$

n is the number of terms in that sequence such that $n>1\in \bf Z$

The possibilities for the sum of the terms of possible sequences are:

$a+(a+1)=2a+1=100$

$a+(a+1)+(a+2)=3a+3=100$

$a+(a+1)+(a+2)+(a+3)=4a+6=100$

...

Or in general: $na+\sum_{i=1}^{n-1}i=100$ $\Rightarrow na+\frac{n(n-1)}{2}=100$

$\Rightarrow n^{2}+(2a-1)n-200=0$

2a-1 is always an odd integer. And since n is also only an integer, this quadric is factorable. To factor this quadratic, there are 6 possible factorizations for 200, and there are two possible sums for each factorization. These sums will give you the possibilities for 2a-1:

1) 1 and 200 $\Rightarrow -1+200=199$ or $1-200=-199$

2) 2 and 100 $\Rightarrow -2+100=98$ or $2-100=-98$

3) 4 and 50 $\Rightarrow -4+50=46$ or $4-50=-46$

4) 5 and 40 $\Rightarrow -5+40=35$ or $5-40=-35$

5) 8 and 25 $\Rightarrow -8+25=17$ or $8-25=-17$

6) 10 and 20 $\Rightarrow -10+20=10$ or $10-20=-10$

Since 2a-1 must be odd, options 2, 3 and 6 are out. You can now use these sums to solve for a:

1) $2a-1=199\Rightarrow a=100$

2) $2a-1=-199\Rightarrow a=-99$

3) $2a-1=35\Rightarrow a=18$

4) $2a-1=-35\Rightarrow a=-17$

5) $2a-1=17\Rightarrow a=9$

6) $2a-1=-17\Rightarrow a=-8$

Since a<100, option 1 is out. By substituting your 5 possible values of 2a-1, you now have 5 quadratics you can solve:

1) $n^{2}-199n-200=0\Rightarrow n=200,-1$

2) $n^{2}+35n-200=0\Rightarrow n=5,-40$

3) $n^{2}-35n-200=0\Rightarrow n=40,-5$

4) $n^{2}+17n-200=0\Rightarrow n=8,-25$

5) $n^{2}-17n-200=0\Rightarrow n=25,-8$

Since n>1, only one of each pair of n values works, leaving you $\color{#D61F06}{5}$ pairs of answers:

1) A sequence of 200 starting with -99

2) A sequence of 5 starting with 18

3) A sequence of 40 starting with -17

4) A sequence of 8 starting with 9

5) A sequence of 25 starting with -8 $\space\space\space\Box$

I start with the fact that $\sum_{k=m+1}^{k = n} k = \frac{n(n+1)}2 - \frac{m(m+1)}2 = 100.$ (It is easy to see this for positive integers, but it works for negatives as well.)

Multiply by two and work out the products:

$n^2-m^2 + n-m = 200; \\ (n-m)(n+m+1) = 200 = 2^3 \cdot 5^2.$

One of the factors will be even, the other odd. Thus one of the factors will be a power of five. (They cannot be negative.) However, we require $n > m$ ). It is now easy to check all cases:

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 n-m  n+m+1   n     m   sequence
----- ----- ----- ----- ------------------
   1   200   100    99  100 (does not count)
   5    40    22    17  18+19+20+21+22
  25     8    16    -9  -8+-7+...+15+16
 200     1   100  -100  -99+-98+...+99+100
  40     5    22   -18  -19+-18+...+21+22
   8    25    16    8   9+10+...+15+16

Thus there are $\boxed{5}$ of such sequences.

100 = 1×100 = 2×50 = 4×25 = 5×20 = 10×10

If there is an odd factor, we can construct a sequence satisfy the condition.

100×1 : a hundred pairs of "1"

(0 + 1) + (-1 + 2) + (-2 + 3) + ... + (-99 + 100) = -99 + -98 + ... + 99 + 100

4×25 : four pairs of "25"

(12 + 13) + (11 + 14) + (10 + 15) + (9 + 16) = 9 + 10 + ... + 15 + 16

25×4 : twenty five "4" (3 + 5 = two "4")

4 + (3 + 5) + (2 + 6) + (1 + 7) + ... + (-8 + 16) = - 8 - 7 + ... + 15 + 16

5×20 : five "20" (19 + 21 = two "20")

20 + (19 + 21) + (18 + 22) = 18 + 19 + 20 + 21 + 22

20×5 : twenty pairs of "5"

(2 + 3) + (1 + 4) + (0 + 5) + ... + (-17 + 22) = - 17 - 16 + ... + 21 + 22