Dear Uncle Colin,

My textbook gives me an arithmetic sequence that starts $3 + 8 + 13 + \dots$, and asks me to find where the sum is 1575.

I’ve got it down to $\frac{-1 \pm \sqrt{63,001}}{10}$, but I don’t know how to work out that square root without a calculator!

- Not Terribly Experienced Resolving Mathematical Summation

Hi, NTERMS, and thanks for your message!

Adding up an arithmetic sequence, as you’ve correctly worked out, involves using the sum formula, which takes two forms: I usually prefer $\frac{1}{2} n (a + L)$, but here I think $\frac{1}{2}n(2a + (n-1)d)$ is probably your friend.

We know that $a=3$ and $d=5$, so we’re solving $1575 = \frac{1}{2}n(6 + 5(n-1))$.

Double everything: $3150 = n + 5n^2$.

Does it factorise? Let’s try the AC method on $5n^2 + n - 3150 = 0$. We’ve got $5 \times 3150 = 15,750$, so we want two factors of that that differ by one. Eyeballing the square root of 15,750, it’s got to be in the region of 120 or 130, and it’s a multiple of 25 - so maybe we should try 125? 15,750 ÷ 125 is 3,150 ÷ 25 or 126 - which works! We’ve got $5n^2 + 126n - 125n - 3150 = 0$, or $n(5n+126) - 25(5n+126) = 0$. We can rewrite it as $(5n+126)(n-25)$ and say “oh look! It must be 25.”

But say you don’t fancy the factorising

Don’t let the Ninja hear you say that! Well, you could always use the formula - the roots turn out to be $\frac{-1 \pm \sqrt{63,001}}{10}$, which leads to a further question: what’s the square root of $63,001$?

Well, $250^2 = 62,500$, so we’re in that sort of ballpark. If we wanted to, we could say $(250+x)^2 = 63,001$, then work it through: $62,500 + 500x + x^2 = 63,001$, so $x^2 + 500x = 501$ and $x=1$ is clearly a solution - so $63,001$ is $251^2$.

We can take that and say the roots are $\frac{-1 \pm 251}{10}$, which gives 25 and a horrible negative fraction as solutions again.

Hope that helps!

- Uncle Colin