How many terms of the progression $3,6,9,12,\dots$ must be taken to have a sum not less than $2000$? How many terms of the progression $3,6,9,12,\dots$ must be taken to have a sum not less than $2000$?
I tried the calculation, say $n$ terms, as they are in AP so are they want this?
${n\over 2}[2\times 3 (n-1)3]\ge 2000$?
but how to find $n$ then?

The progression $3,6,9,12, \ldots $ is a sequence $(a_n)_{n \in \mathbb{N}}$, $a_n = 3n$. The sum of this sequence can be written as $$\sum\limits_{i=1}^{n} 3i = 3\cdot\sum_{i=1}^{n}i$$
And we have the sum of a sequence formula that states $\sum\limits_{i=1}^{n}i = \frac{n(n+1)}{2}$. It follows that $$\sum_{i=1}^{n}3i = 3\left(\frac{n(n+1)}{2}\right) = \frac{3n^2+3n}{2}$$.
Now, you want $\frac{3n^2+3n}{2} \geq 2000$. So we'll find the value of $n$ such that $\frac{3n^2+3n}{2} =2000$. Any larger value of $n$ will obviously give you a value greater than $2000$.
$$\frac{3n^2+3n}{2} = 2000$$
$$\Rightarrow 1.5n^2+1.5n=2000$$
$$1.5n^2+1.5n-2000=0$$
Solve the quadratic equation.
$$n = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$n = \frac{-1.5\pm\sqrt{2.25- (-12000)}}{3}$$
$$n = \frac{-1.5\pm 109.5547808}{3}$$
So $n=36.02\ldots$ or $n=-37.02\ldots$. Discard the negative value, as you can't have a series with a negative number of terms. We also need to round up to get an integer term, and we can't round down because we found the minimum $n$ to be $36+\epsilon$, so $36<n$, so $36$ will not produce a satisfactory sum. So $n=37$. We can test it:
$$\frac{3(37^2)+3(37)}{2} = 2109$$
$$\frac{3(36^2)+3(36)}{2} = 1998$$
And that confirms that $37$ is indeed the minimum term to have a satisfactory sum: our progression must have at least $37$ terms. (I.e. our progression will be $3(1),3(2),\ldots,3(37)$, which is $3,6,9,12,\ldots 111$.)
