Find all possible ways of expressing as the sum of consecutive natural numbers.

Understand the Question Let be written as the sum of consecutive natural numbers starting from , where and : Using the formula for the sum of an arithmetic progression: Since the sum of and is (an odd number), one factor must be odd and the other even. Additionally, since , we have . We can find all valid combinations by testing the odd factors of . Step 1 · Form the equation for the sum of consecutive numbers Let the consecutive natural numbers be , where and . Sum of terms in AP Step 2 · Find valid factor pairs of 200 Since with , and the two factors must have opposite parities (one odd, one even), we find the odd factors of . Prime factorisation of The odd factors of are and . Case 1: Odd factor The consecutive numbers are Case 2: Odd factor The consecutive numbers are Answer > There are 2 ways: 1. 2. Common Mistakes Including : When , the expression is just , which is a single number and not a sum of consecutive natural numbers. Parity Mismatch: Trying factor pairs where both factors are even (e.g., ), which leads to non-integer values for since and must have opposite parity.

Predicting What Comes Next: Sequences and Progressions | EOT

Question 6

Find all possible ways of expressing $100$ as the sum of consecutive natural numbers.

Solution

Understand the Question

Let $100$ be written as the sum of $k$ consecutive natural numbers starting from $a$ , where $a \ge 1$ and $k \ge 2$ :

a + (a + 1) + (a + 2) + \dots + (a + k - 1) = 100

Using the formula for the sum of an arithmetic progression:

\text{Sum} = \dfrac{k}{2}[2a + (k - 1)] = 100 \implies k(2a + k - 1) = 200

Since the sum of $k$ and $(2a + k - 1)$ is $2a + 2k - 1$ (an odd number), one factor must be odd and the other even.
Additionally, since $a \ge 1$ , we have $2a + k - 1 > k$ . We can find all valid combinations by testing the odd factors of $200$ .

Step 1 · Form the equation for the sum of consecutive numbers

Let the consecutive natural numbers be $a, a+1, a+2, \dots, a+k-1$ , where $a \ge 1$ and $k \ge 2$ .

Sum of $k$ terms in AP

S_k = \dfrac{k}{2}[2a + (k - 1)] = 100

k(2a + k - 1) = 200

Step 2 · Find valid factor pairs of 200

Since $k(2a + k - 1) = 200$ with $2a + k - 1 > k$ , and the two factors must have opposite parities (one odd, one even), we find the odd factors of $200$ .

Prime factorisation of $200$

200 = 2^3 \times 5^2

The odd factors of $200$ are $1, 5,$ and $25$ .

Case 1: Odd factor $= 5$

k = 5, \quad 2a + k - 1 = 40

2a + 5 - 1 = 40 \implies 2a = 36 \implies a = 18

The $5$ consecutive numbers are

18, 19, 20, 21, 22

Case 2: Odd factor $= 25$

2a + k - 1 = 25, \quad k = 8

2a + 8 - 1 = 25 \implies 2a = 18 \implies a = 9

The $8$ consecutive numbers are

9, 10, 11, 12, 13, 14, 15, 16

Answer

There are 2 ways:

$18 + 19 + 20 + 21 + 22 = 100$
$9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 100$

Common Mistakes

Including $k = 1$ : When $k = 1$ , the expression is just $100$ , which is a single number and not a sum of consecutive natural numbers.
Parity Mismatch: Trying factor pairs where both factors are even (e.g., $10 \times 20 = 200$ ), which leads to non-integer values for $a$ since $k$ and $(2a + k - 1)$ must have opposite parity.

Question 6

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