Sum of Consecutive Odd Integers Starting with 5 to Reach 480

Let us explore the problem of determining the number of consecutive odd integers, beginning with 5, that sum to 480. We will employ the properties of arithmetic progressions (AP) to solve this intriguing mathematical problem.

Solving the Problem Using Arithmetic Progression (AP)

We start with the first term a 5 and the common difference d 2. The general formula for the sum of the first n terms of an AP is given by:

S_n frac{n}{2} [ 2a ( n - 1 )d ]

Substituting the values of a 5 and d 2 into the formula, we get:

S_n frac{n}{2} [ 2(5) ( n - 1 )2 ] frac{n}{2} [ 10 2n - 2 ] frac{n}{2} [ 8 2n ] n(4 n)

The sum equals 480, so we have:

n(4 n) 480

Rearranging and simplifying, we get the quadratic equation:

n^2 4n - 480 0

Factoring this quadratic equation, we get:

n^2 24n - 20n - 480 0

n(n 24) - 20(n 24) 0

(n - 20)(n 24) 0

Therefore, the solutions are n 20 and n -24. Since n cannot be negative, we take n 20.

Verification and Conclusion

The 20th term of the sequence can be calculated as:

T_{20} a ( 20 - 1 )d 5 19(2) 5 38 43

Hence, the sequence is:

5, 7, 9, 11, ..., 43

This sequence comprises 20 terms, and the sum is 480.

Additional Insight

Now, consider the scenario where the problem refers to consecutive two-digit odd integers starting with 5. We assume these numbers to be of the form 5, 15, 25, ..., 51.

The first term is a 51 and the common difference is d 2. The n-th term of the AP can be written as:

T_n a ( n - 1 )d 51 ( n - 1 )2 49 2n

The sum of the first n terms is:

S_n frac{n}{2} [ 2a ( n - 1 )d ] frac{n}{2} [ 2(51) ( n - 1 )2 ] frac{n}{2} [ 102 2n - 2 ] frac{n}{2} [ 100 2n ] 50n n^2

Solving for S_n 480, we get:

50n n^2 480

n^2 50n - 480 0

Approximatively solving this quadratic equation, we get:

n approx 8.24154

Summing the terms for n 8 and n 9 gives approximately:

For n 8, S_8 464

For n 9, S_9 531

Hence, there is no integer n that makes the sum exactly 480 in this scenario.

Lastly, for further insight, we recognize that the sum of the first n odd numbers is a perfect square:

S_n 1 3 5 ... ( 2n - 1 ) n^2

Checking if adding 1 and 3 to 480 makes it a perfect square, we get:

1 3 480 484 22^2

Hence, this sum can be represented as:

2n - 1 2(22) - 1 43

Therefore, n 43.

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