Finding the Sum of All Two-Digit Odd Numbers

The problem asks for the sum of all positive odd numbers that have two digits. First, let's identify these numbers and the type of sequence they form.

Identifying the Sequence of Two-Digit Odd Numbers

Two-digit numbers range from 10 to 99. The odd numbers in this range are:

11, 13, 15, 17, ..., 97, 99

This sequence is an arithmetic progression (AP) because the difference between consecutive terms is constant.

• First term (\(a\)): The smallest two-digit odd number is 11.
• Last term (\(l\)): The largest two-digit odd number is 99.
• Common difference (\(d\)): The difference between consecutive odd numbers is 2 (\(13 - 11 = 2\), \(15 - 13 = 2\), and so on).

Calculating the Number of Terms

To find the sum of an arithmetic progression, we need to know the number of terms (\(n\)) in the sequence. We can use the formula for the \(n\)-th term of an AP:

\(l = a + (n-1)d\)

Substitute the values we know:

\(99 = 11 + (n-1)2\)

Now, solve for \(n\):

• Subtract 11 from both sides: \(99 - 11 = (n-1)2\)
• \(88 = (n-1)2\)
• Divide both sides by 2: \(\frac{88}{2} = n-1\)
• \(44 = n-1\)
• Add 1 to both sides: \(n = 44 + 1\)
• \(n = 45\)

So, there are 45 two-digit odd numbers.

Calculating the Sum of the Arithmetic Progression

The sum of an arithmetic progression (\(S_n\)) can be calculated using the formula:

\(S_n = \frac{n}{2} (a + l)\)

We have the number of terms (\(n = 45\)), the first term (\(a = 11\)), and the last term (\(l = 99\)). Substitute these values into the formula:

\(S_{45} = \frac{45}{2} (11 + 99)\)

Perform the calculation:

• Add the terms in the parenthesis: \(11 + 99 = 110\)
• \(S_{45} = \frac{45}{2} (110)\)
• Multiply \(\frac{45}{2}\) by 110: \(S_{45} = 45 \times \frac{110}{2}\)
• \(S_{45} = 45 \times 55\)

Now, calculate \(45 \times 55\):

\(45 \times 55 = 45 \times (50 + 5)\)

\(= 45 \times 50 + 45 \times 5\)

\(= 2250 + 225\)

\(= 2475\)

The sum of all two-digit odd numbers is 2475.

Summary of the Calculation

Here's a quick recap of the steps:

• Identified the sequence: 11, 13, ..., 99.
• Determined it's an AP with \(a = 11\), \(l = 99\), \(d = 2\).
• Calculated the number of terms \(n = 45\).
• Applied the sum formula \(S_n = \frac{n}{2} (a + l)\).
• Calculated \(S_{45} = \frac{45}{2} (11 + 99) = \frac{45}{2} \times 110 = 45 \times 55 = 2475\).

The sum of all two digit odd numbers is 2475.

Revision Table: Sum of Two-Digit Odd Numbers

Additional Information: Arithmetic Progressions and Sums

An arithmetic progression (AP) is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference (\(d\)).

Key formulas for an AP:

• \(n\)-th term (\(a_n\)): \(a_n = a + (n-1)d\), where \(a\) is the first term.
• Sum of the first \(n\) terms (\(S_n\)):
  • If you know the last term (\(l\)): \(S_n = \frac{n}{2} (a + l)\)
  • If you don't know the last term: \(S_n = \frac{n}{2} (2a + (n-1)d)\)

In this problem, identifying the two-digit odd numbers as an AP was crucial. The first term is 11, the last term is 99, and the common difference is 2. Using these values, we could easily find the number of terms and then apply the sum formula.

This method is generally applicable to finding the sum of any arithmetic sequence, whether it's odd numbers, even numbers, or any other sequence with a constant difference between terms.