The Beauty of Arithmetic Progressions
Imagine you are standing at the start of a sequence of numbers. You know that each step you take, you add the same amount to get to the next number. This is the elegance of an Arithmetic Progression (A.P.).
In this problem, we are dealing with a sequence of positive integers. This is a crucial piece of information; it is not just flavor text, but a mathematical constraint that acts as our compass, guiding us through the algebraic landscape.
Decoding the Constraints
We begin by defining our A.P. with a first term a and a common difference d. The problem states that all terms are positive integers.
This implies that both a and d must be integers. If a is an integer and a+d is an integer, then d must also be an integer. Keep this in mind; it is the key that will unlock the final answer.
The S3 Reduction
We are given that the sum of the first three terms is 54. Using the sum formula Sn=2n[2a+(n−1)d], we can write:
Simplifying this, we get 3(a+d)=54, which leads us to the relation:
This means we can express our first term as a=18−d. We have successfully reduced our problem from two variables to one.
The S20 Inequality
Now, let us look at the sum of the first twenty terms, S20. Using the formula again:
S20=220[2a+19d]=10[2a+19d]
By substituting our expression a=18−d into this, we get:
S20=10[2(18−d)+19d]=10[36−2d+19d]=10[36+17d]
The problem states that 1600<S20<1800. Substituting our expression for S20, we get the inequality:
Dividing by 10, we have 160<36+17d<180. Subtracting 36 from all parts, we get:
The Final Key
Dividing by 17, we find that 7.29<d<8.47. Since we established that d must be an integer, the only possible value for d is 8.
With d=8, we can easily find a=18−8=10. Finally, we calculate the 11th term using the formula Tn=a+(n−1)d:
The journey from the initial constraints to the final result is a testament to how algebraic manipulation, when guided by logical constraints, leads us directly to the truth. The final answer is 90.