Question 6

Let's find the first number $a$ and the number of terms $n$ for the sum of consecutive natural numbers.

Step 1 — Set up the equation

We represent the sum of $n$ consecutive natural numbers starting from $a$. The sum of an arithmetic progression is given by the formula.

$S_n = \frac{n}{2}(2a + (n-1)d)$

Here, the common difference $d$ is 1. The sum $S_n$ is 100.

$100 = \frac{n}{2}(2a + n - 1)$

Let's multiply both sides by 2.

$200 = n(2a + n - 1)$

We need to find integer values for $a$ and $n$. From this equation, $n$ must be a factor of 200. Also, we can write $2a$ as:

$2a = \frac{200}{n} - n + 1$

Since $a$ is a natural number, $a$ must be greater than or equal to 1. So, $2a$ must be greater than or equal to 2.

$\frac{200}{n} - n + 1 \ge 2$

Let's simplify this inequality.

$\frac{200}{n} - n - 1 \ge 0$

$\frac{200 - n^2 - n}{n} \ge 0$

Since $n$ is the number of terms, $n$ must be a positive integer. So, the numerator must be non-negative.

$200 - n^2 - n \ge 0$

$n^2 + n - 200 \le 0$

Let's find the roots of the quadratic equation $n^2 + n - 200 = 0$.

$n = \frac{-1 \pm \sqrt{1^2 - 4(1)(-200)}}{2(1)}$

$n = \frac{-1 \pm \sqrt{1 + 800}}{2}$

$n = \frac{-1 \pm \sqrt{801}}{2}$

The approximate value of $\sqrt{801}$ is 28.3.

$n \approx \frac{-1 \pm 28.3}{2}$

Since $n$ must be positive, we take the positive root.

$n \approx \frac{-1 + 28.3}{2}$

$n \approx \frac{27.3}{2}$

$\boxed{n \approx 13.65}$

This means $n$ must be an integer less than or equal to 13.

Step 2 — Identify possible values for n

We know $n$ must be a factor of 200. We also know $n \le 13$. Let's list the factors of 200: 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 200. The factors of 200 that are less than or equal to 13 are:

$\boxed{n \in \{1, 2, 4, 5, 8, 10\}}$

Step 3 — Check each possible value of n

We use the equation $2a = \frac{200}{n} - n + 1$. For $a$ to be a natural number, $2a$ must be an even integer. This means $\frac{200}{n} - n + 1$ must be an even integer. Equivalently, $\frac{200}{n} - n$ must be an odd integer.

Case 1: $n = 1$ Let's substitute $n = 1$ into the equation for $2a$.

$2a = \frac{200}{1} - 1 + 1$

$2a = 200$

$a = 100$

This gives the sum: 100.

$100 = 100$

Case 2: $n = 2$ Let's substitute $n = 2$ into the equation for $2a$.

$2a = \frac{200}{2} - 2 + 1$

$2a = 100 - 2 + 1$

$2a = 99$

Since 99 is odd, $a$ is not an integer. This case is not valid.

Case 3: $n = 4$ Let's substitute $n = 4$ into the equation for $2a$.

$2a = \frac{200}{4} - 4 + 1$

$2a = 50 - 4 + 1$

$2a = 47$

Since 47 is odd, $a$ is not an integer. This case is not valid.

Case 4: $n = 5$ Let's substitute $n = 5$ into the equation for $2a$.

$2a = \frac{200}{5} - 5 + 1$

$2a = 40 - 5 + 1$

$2a = 36$

$a = 18$

This gives the sum of 5 terms starting from 18. The terms are 18, 19, 20, 21, 22.

$100 = 18 + 19 + 20 + 21 + 22$

Case 5: $n = 8$ Let's substitute $n = 8$ into the equation for $2a$.

$2a = \frac{200}{8} - 8 + 1$

$2a = 25 - 8 + 1$

$2a = 18$

$a = 9$

This gives the sum of 8 terms starting from 9. The terms are 9, 10, 11, 12, 13, 14, 15, 16.

$100 = 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16$

Case 6: $n = 10$ Let's substitute $n = 10$ into the equation for $2a$.

$2a = \frac{200}{10} - 10 + 1$

$2a = 20 - 10 + 1$

$2a = 11$

Since 11 is odd, $a$ is not an integer. This case is not valid.

We have found all possible ways.

Answer

(i) $100 = 100$ (ii) $100 = 18 + 19 + 20 + 21 + 22$ (iii) $100 = 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16$