Answer :
Final answer:
The numbers whose sum is 15 and are in an arithmetic progression (AP) will still be in an AP when 1419 is added to them. Option 1 is the correct answer.
Explanation:
Let's assume the three numbers forming the arithmetic progression are a−d, a, and a+d, where d is the common difference. The sum of these numbers is 3a. Given that the sum is 15, we find 3a=15, leading to a=5. Now, the original three numbers are 5−d, 5, and 5+d. When 1419 is added to each: (5−d)+1419=1424−d, 145+1419=1424, and (5+d)+1419=1424+d. The resulting numbers are 1424−d, 1424, and 1424+d.
The differences between consecutive terms remain constant, indicating an arithmetic progression. the numbers, initially forming an arithmetic progression with a sum of 15, transition into a new arithmetic progression when 1419 is added to each term. The resulting sequence, with terms 1424−d, 1424, and 1424+d, maintains a constant difference between successive elements, signifying the preservation of an arithmetic progression.
This analysis underscores the transformative impact of the addition of 1419 on the nature of the sequence, reinforcing its adherence to the arithmetic progression pattern.
