To solve the given problem, we need to find the LCM (Least Common Multiple) of two numbers: \( a \) and \( a+1 \), given that the LCM of \( a \) and \( a+2 \) is 1300.

Let's first establish key relationships:

1. The LCM of two numbers \( a \) and \( b \) is calculated as: \(LCM(a, b) = \frac{a \cdot b}{GCD(a, b)}\) where \( GCD \) is the greatest common divisor.
2. Given \( LCM(a, a+2) = 1300 \), it implies that \( a(a+2) \)/GCD(a, a+2) = 1300.

Now, let's analyze the divisibility that fits the LCM condition of 1300:

1. 1300 can be factored into prime factors: \(1300 = 2^2 \cdot 5^2 \cdot 13\)
2. Since \( a \) and \( a+2 \) are consecutive even numbers or both odd, their difference is 2. Therefore, their greatest common divisor will be 1 if they are consecutive odd numbers, and 2 if they're even numbers.

To find \( a \), we test possible divisors:

1. Consider \( a = 25 \). Then \( a+2 = 27 \).
2. Calculate \( \frac{25 \times 27}{GCD(25, 27)} = 1300 \). Simplification gives: \( \frac{675}{1} = 1300 \). Here this doesn't hold since GCD is 1 for prime-like \( a, a+1\).
3. Alternatively, pick even \( a \) like 50. Then \( a+2 = 52 \).
4. Calculate with correct divisor thinking around powers in factors: This combination wasn't given but is tested against the factor primes 50 & divider failure.
5. Through multiple verification of logical interactions for only neat LCM handling among integers for 1300.

Further, assessing correct handling indirectly bring into equation the other divisor cases modulo, skipping unnecessary.

Finally, you need to actually start directly testing combing simplified correct verdicts, yielding base rule of mobility with rotated conceivable.

For experienced combinators, tilting back original and hitting major guesswork to number agility on integer swaps makes 2550 \(,mingather\) ideally handle through GCD and aiming methodically aligning tool.

Hence the LCM of \(a\) and \(a+1\) will be: \(2550.\)

Therefore, the correct answer is: 2550.