The number of terms common to the two A.P.'s 3, 7, 11, ....., 407 and 2, 9, 16, ....., 709 is ______.

Solution

Maths sir · 4 steps

Common terms form their own A.P. Find its first term, step by LCM(4, 7) = 28, and stop at 407.

1. First common term. List the first A.P. as 3, 7, 11, 15, 19, 23 and the second as 2, 9, 16, 23. The first number in both lists is 23.
2. Common difference. After 23, a number stays in both A.P.s only if you add a multiple of 4 and a multiple of 7 at the same time. The smallest such jump is LCM(4, 7) = 28. So the common terms are 23, 51, 79, and so on.
3. Upper limit. The first A.P. ends at 407, so no common term can be larger than 407, even though the second A.P. goes on to 709.

   23 + (n-1)× 28 ≤ 407 ⇒ n - 1 ≤ 384/28 ≈ 13.7

4. So. The largest whole value is n - 1 = 13, so n = 14. The last common term is 23 + 13 × 28 = 387, and there are 14 common terms.

The common slip: 13. Solving 28(n-1) ≤ 384 and taking n = 13 forgets the +1 for the first term 23, so it gives 13, not 14.

The idea to keep. Terms common to two A.P.s form a new A.P. Its difference is the LCM of the two differences, and its last term cannot pass the smaller of the two last terms.

Answer from the NTA final answer key. Solution by PrepWiser.