The first three terms of an arithmetic series are $-\frac{9}{8}, -\frac{5}8, -\frac{1}8$.
The sum $S_{\text{n}}$ of the first n terms of this series is $\frac {n}{8}(2n-11)$, for n = 1, 2, 3, ...
Find the value of n when $S_{\text{n}}$ is the smallest.
Answer given:
$S_{\text{n}}$ is symmetrical about $x=\frac{11}4$ : $S_{\text{n}}$ is smallest when n=3.
My question is what does it mean to be symmetrical about $x$? And doesn't the smallest of the sum of the series is when $S_{\text{1}}=T_{\text{1}}$?
What does it mean to be symmetrical about $\dfrac{11}4$?
It means if $f(n)=\dfrac n8(2n-11)$, then $f\left(\dfrac{11}4-x\right)=f\left(\dfrac{11}4+x\right).$
This may be easier to see if you complete the square: $f(n)=\left(\dfrac n2 - \dfrac{11}8\right)^2-\dfrac{121}{64}$.
$f(x)$ is minimal when $x$ is $\dfrac{11}4$, so $f(n)$ is minimal when $n=3, $ which is the integer closest to $\dfrac{11}4$.
And this is apparent from
$S_1=-\dfrac98, S_2=-\dfrac{14}8, S_3=-\dfrac{15}8, S_4=-\dfrac{12}8, S_5=-\dfrac58, S_6=\dfrac68, S_7=\dfrac{21}8, ...$