Introduction to Arithmetic Progressions

Arithmetic progressions (AP) are a fundamental concept in mathematics, where each term after the first is obtained by adding a constant difference to the previous term. Understanding how to solve for the smallest possible value of $n$ when the sum of the first $n$ terms of an AP is greater than 2000 is both challenging and valuable for students and professionals in mathematics and related fields.

Understanding the Problem

Given an arithmetic progression with the first term $a_1 6$ and the fifth term $a_5 18$, the goal is to determine the smallest number of terms $n$ such that the sum of the first $n$ terms exceeds 2000.

Solving for the Common Difference

The common difference $d$ in an arithmetic progression can be calculated using the formula:

$a_n a_{n-1} d$

Given:

$a_1 6$ $a_5 18$

The formula for the fifth term is:

$a_5 a_1 4d$

Substitute the known values:

$18 6 4d$

By solving for $d$:

$12 4d$

$d 3$

General Term Formula

The general term $a_n$ for an arithmetic progression can be expressed as:

$a_n a_1 (n-1)d$

Substitute $a_1 6$ and $d 3$:

$a_n 6 (n-1)3 3n 3$

Sum of the First n Terms

The sum of the first $n$ terms of an arithmetic progression, $S_n$, is given by:

$S_n frac{n}{2} cdot (a_1 a_n)$

Substitute $a_1 6$ and $a_n 3n 3$:

$S_n frac{n}{2} cdot (6 (3n 3)) frac{n}{2} cdot (6 3n 3) frac{n}{2} cdot (3n 9) frac{3n^2 9n}{2}$

Setting Up the Inequality

We need to find the smallest $n$ such that:

$S_n > 2000$

Substitute the sum formula:

$frac{3n^2 9n}{2} > 2000$

Multiply both sides by 2:

$3n^2 9n > 4000$

Rearrange:

$3n^2 9n - 4000 > 0$

Divide the whole inequality by 3:

$n^2 3n - frac{4000}{3} > 0$

Rationalize the constant:

$n^2 3n - 1333.33 > 0$

Solve the quadratic equation:

Using the quadratic formula $n frac{-b pm sqrt{b^2 - 4ac}}{2a}$, where $a 1$, $b 3$, and $c -1333.33$:

$n frac{-3 pm sqrt{3^2 - 4 cdot 1 cdot -1333.33}}{2 cdot 1}$

$n frac{-3 pm sqrt{9 5333.32}}{2}$

$n frac{-3 pm sqrt{5342.32}}{2}$

$n frac{-3 pm 73.2}{2}$

Two solutions:

$n approx frac{-3 73.2}{2} approx 35.1$ $n approx frac{-3 - 73.2}{2}$ (negative, not applicable)

Rounding up, the smallest possible value of $n$ is 36.

Conclusion

Hence, the smallest possible value of $n$ such that the sum of the first $n$ terms is greater than 2000 is:

boxed{36}