The question "How many terms are common to the arithmetic progression 3 7 11?" is incomplete. It only provides the beginning of an arithmetic progression (3, 7, 11, ...). To determine the number of common terms, we need at least two arithmetic progressions to compare. Furthermore, we need to know how many terms are in each progression, or a stopping point for each. Without this information, the number of common terms cannot be determined.
However, if the question were "Given two arithmetic progressions, AP1 and AP2, how many terms are common to both? AP1 is given by 3, 7, 11,... and has its nth term less than or equal to 60. AP2 is given by 2, 5, 8,... and has its nth term less than or equal to 70." We can solve.
Example with Two Arithmetic Progressions
Let's complete the problem:
AP1: 3, 7, 11, ... with nth term ≤ 60.
AP2: 2, 5, 8, ... with nth term ≤ 70.
We are given that for some n:
- AP1n ≤ 60
- AP2n ≤ 70
Finding the Number of Terms in Each AP
AP1: 3, 7, 11,...
The general term of AP1 is given by: an = a1 + (n-1)d, where a1 = 3 and d = 4. Therefore:
an = 3 + (n - 1)4
an = 3 + 4n - 4
an = 4n - 1
We want to find n such that 4n - 1 ≤ 60:
4n - 1 ≤ 60
4n ≤ 61
n ≤ 61/4
n ≤ 15.25
Since n must be an integer, the largest possible value for n is 15. Therefore, AP1 has 15 terms that are less than or equal to 60.
AP2: 2, 5, 8,...
The general term of AP2 is given by: an = a1 + (n-1)d, where a1 = 2 and d = 3. Therefore:
an = 2 + (n - 1)3
an = 2 + 3n - 3
an = 3n - 1
We want to find n such that 3n - 1 ≤ 70:
3n - 1 ≤ 70
3n ≤ 71
n ≤ 71/3
n ≤ 23.666...
Since n must be an integer, the largest possible value for n is 23. Therefore, AP2 has 23 terms that are less than or equal to 70.
Finding Common Terms
AP1 = {3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59}
AP2 = {2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68}
The common terms are: {11, 23, 35, 47, 59}. There are 5 common terms.
Alternate example using the prompt's reference
If instead AP1 consists only of terms less than or equal to 60, and AP2 consists only of the terms less than or equal to 70 from another progression such that the nth term of common difference 5 is less than or equal to 70. The reference provided "⇒n≤14.7⇒n=14" implies that 14 is relevant somehow.
Let's assume that AP2 has the form 1, 6, 11, 16, ... The common difference here is 5. Then the nth term can be written as:
an = 1 + (n-1)5
If we want to find n such that an ≤ 70:
1 + (n-1)5 ≤ 70
1 + 5n - 5 ≤ 70
5n - 4 ≤ 70
5n ≤ 74
n ≤ 14.8
Then, n = 14. The terms for this new AP2 are: {1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, 61, 66}
AP1 = {3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59}
Common terms: {11, 31, 51}. So there are 3 common terms.
