How many integers appear in both of the following arithmetic progressions:
A1: 2, 9, 16,.....2+(1000-1)*7
A2: 3, 12, 21,.....3+(1000-1)*9
Since 2 appears in A1 but not in A2, it does not appear in both of the arithmetic progressions.
A1_n= 2+(n-1)*7=7n-5
A2_k= 3+(k-1)*9=9k-6
Put A1_n=A2_k => so 9k-7n=1 and solve the Diophantine equation for n,k u can see that gcd(9,7)=1 so the solutions you want come from applying the euclidean algorithm and your numbers n,k must be less than or equal to 1000.
Calculating GCD(9,7) gives:
7 = 0*9 + 7
9 = 1*7 + 2
7 = 3*2 + 1
2 = 2*1 + 0
Then applying the Extended Euclidean Algorithm:
1 = (1 * 7) + (-3 * 2)
= (-3 * 9) + (4 * 7)
= (4 * 7) + (-3 * 9)
A particular solution is:
x0 = -3
y0 = 4
The complete solution is:
x = -3 + 7n
y = 4 - 9n
To find the integers that appear in both arithmetic progressions, we need to identify the terms that are common to both A1 and A2.
Let's analyze the pattern of A1 first.
A1: 2, 9, 16, ...
From the given pattern, we can observe that the difference between consecutive terms is 7. This indicates that each term is obtained by adding 7 to the previous term.
The general formula to find the nth term of an arithmetic progression is:
An = A1 + (n-1)d
In this case, A1 = 2 (the first term), and d = 7 (the common difference).
To find the last term of A1, we substitute the values into the formula:
A1 + (n-1)d = 2 + (1000-1)*7
Simplifying further, we have:
A1 + 6d = 2 + 6(7)
A1 + 6d = 2 + 42
A1 + 6d = 44
Now, let's analyze the pattern of A2.
A2: 3, 12, 21, ...
Similarly, we observe that the difference between consecutive terms in A2 is 9. Therefore, each term is obtained by adding 9 to the previous term.
Using the same formula, we can find the last term of A2:
A1 + (n-1)d = 3 + (1000-1)*9
Simplifying further:
A1 + 8d = 3 + 8(9)
A1 + 8d = 3 + 72
A1 + 8d = 75
To identify the integers that are common to both A1 and A2, we need to determine the terms that satisfy both equations A1 + 6d = 44 and A1 + 8d = 75.
Subtracting the first equation from the second equation, we get:
2d = 31
Since the common difference (d) should be a whole number, it follows that there is no whole number solution for 2d = 31. Therefore, there are no integers that appear in both A1 and A2.