# 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?

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Expand and use quadratic equation –  E.O. Dec 15 '13 at 2:42
How did you find that formula for the sum? This should be $\sum_{k=0}^n 3k = 3 \frac{n(n+1)}{2}$, right? –  nik Dec 15 '13 at 2:42

The right formula is - as Nik commented, $$S(n):=3+6+9+12+\dots+3n = 3\cdot\frac{n(n+1)}2$$ which is approximately $\frac32\cdot n^2$. So, this should be around $2000$, that is $n^2$ should be around $2/3\cdot 2000\approx 1334$. Using calculator, its square root is $\approx 36.52$ and substituting $n=36$ to $S(n)$ gives $S(36)=1998$, so at least $\bf 37$ terms are needed to exceed $2000$.

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Note you can easily complete the square and use $(n+\frac 12)^2$ instead of $n^2$ - not that it makes much difference, though it makes it clear that $36$ will be too small. –  Mark Bennet Dec 15 '13 at 3:17
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 $3,6,9,12,\ldots 111$.)