In an arithmetic series find the least number of terms k that must be taken so that Sk > 559

Question

$T_{10}=37,S_5=45$ therefore $a=1$ and $d=4.$

Trying to find $k$:

$S_n= \frac{n}{2}(2(1) + (n-1)4 ) > 559$

$4n^2 - 2n > 1118$

How do I finish this inequality to find the least number of n?

Answer 1

$$4n^2-2n>1118$$

$$4n^2-2n+\frac{1}{4}>1118+\frac{1}{4}$$

$$(2n-\frac{1}{2})^2>1118+\frac{1}{4}$$

$33^2=1089$, $34^2=1156$

$$2n-\frac{1}{2}>33.4$$

$$2n>33.9$$

$$n>16.95$$

i.e. $$n=17$$

Answer 2

For finding roots it might be handy As you already reached till the quadratic equation, the next part is to find the roots after solving the equation, we get roots as 16.97 and -16.47

so expression will be (k-16.97)(k+16.47)>0

that indicates us k should be either greater than 16.97 or less than -16.47, as k is the number of terms it can't be negative and the least integer greater than 16.97 would be 17.

So you need to take 17 terms so as to make the sum greater than 559 And if you check, $S_{17}$=561 which satisfies the condition.