# How to prove the common difference of two arithmetic progression is equal to the lcm of both the progressions' differences?

If the first ap's difference is 3, and 2nd's 4, I tried putting it in 3n and 4n terms, and ultimately showing they both will have common roots where 3 and 4 both have terms, as they're coprime. But combining both the progressions proved to be difficult.

2. Could it be done by graph? Taking functions of both the progression as again f(x) = 3n, and f(y) = 4n? On the same graphs, plotting all the terms like this will give some common terms in which both 3n and 4n will be coordinates. I've no idea if this is even possible, it's just an idea.

If there's a way, please enlighten. Thank you.

• Erdos
+1

Your question is vague. What do you mean by "common difference of two arithmetic progression". What do you mean by lcm of both progressions differences?

There are different methods to solve this problem, but one possible way is to use the Chinese Remainder Theorem (CRT). This theorem states that if we have a system of linear congruences, such as:

x ≡ a1 (mod d1)
x ≡ a2 (mod d2)

where a1, a2, d1, and d2 are given integers, then there exists a unique solution x modulo d1 * d2, provided that d1 and d2 are coprime (i.e., they have no common factors other than 1).

To apply this theorem to your problem, you need to write the general terms of the two arithmetic progressions as:

an = a1 + (n - 1) * d1
bn = a2 + (n - 1) * d2

where a1 and a2 are the first terms, and d1 and d2 are the common differences. Then, you need to find the values of n and m such that an = bn, or equivalently:

a1 + (n - 1) * d1 ≡ a2 + (m - 1) * d2 (mod d1 * d2)

This is equivalent to:

(n - m) * d2 ≡ a2 - a1 (mod d1)
(n - m) * d1 ≡ a1 - a2 (mod d2)

These are two linear congruences that can be solved using the CRT.

Once you find the values of n and m that satisfy these congruences, you can plug them into the formulas for an and bn to get the first common term of the two arithmetic progressions.

For example, let's say you have two arithmetic progressions with:

a1 = 5, d1 = 3
a2 = 7, d2 = 4

Then, you need to solve:

(n - m) * 4 ≡ 7 - 5 (mod 3)
(n - m) * 3 ≡ 5 - 7 (mod 4)

Using the CRT, you can find that:

n ≡ 5 (mod 12)
m ≡ 8 (mod 12)

This means that n and m can be any integers that have a remainder of 5 and 8 when divided by 12, respectively. For example, n = 17 and m = 20 are possible values. Then, you can plug them into the formulas for an and bn to get:

a17 = 5 + (17 - 1) * 3 = 53
b20 = 7 + (20 - 1) * 4 = 83

These are not equal, so they are not the first common term. However, if you add 12 to both n and m, you get:

a29 = 5 + (29 - 1) * 3 = 89
b32 = 7 + (32 - 1) * 4 = 127

These are also not equal, but if you add another 12 to both n and m, you get:

a41 = 5 + (41 - 1) * 3 = 125
b44 = 7 + (44 - 1) * 4 = 175

These are equal, so they are the first common term of the two arithmetic progressions. Therefore, x = an = bn = **125** is the answer.

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