Question

Let $d$ be the minimum value of $f\left(x\right)=5x^2-2x+\frac{26}{5}$ and $f\left(x\right)$ is symmetric about $x=r$. If ${\sum }_{n=1}^{\infty }(1+(n-1\left)d\right)r^{n-1}$ equals $\frac{p}{q}$, where $p$ and $q$ are relative prime, then find the value of $(3q-p)$.

Answer 1

Given, $f\left(x\right)=5x^2-2x+\frac{26}{5}$ symmetric about $x=r$

$f\left(x\right)=5x^2-2x+\frac{1}{5}+5=5(x-\frac{1}{5}{)}^2+5$

$\Rightarrow f\left(x\right)$ is symmetric about $x=\frac{1}{5}$ with min values, $d=5$

Now, $S=\sum _{n=1}^{\infty }(1+(n-1\left)d\right)r^{n-1}=\sum _{n=1}^{\infty }[1+(n-1\left)5\right](\frac{1}{5}{)}^{n-1}$

$S=1+\frac{6}{5}+\frac{11}{5^2}+\frac{16}{5^3}+....$ [A.G.P series]

$\frac{1}{5}.S=\frac{1}{5}+\frac{6}{5^2}+\frac{11}{5^3}+...$

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$\frac{4}{5}S=1+\frac{5}{5}+\frac{5}{5^2}+\frac{5}{5^3}+....$

$=2+(\frac{1}{5}+\frac{1}{5^2}+....)$

$=2+\frac{\frac{1}{5}}{1-\frac{1}{5}}$ [sum of infinite G.P = $\frac{a}{1-r}]$

$\Rightarrow \frac{4}{5}S=2+\frac{1}{4}=\frac{9}{4}$

$S=\frac{45}{16}\Rightarrow P=45,q=16$

$3q-p=3\left(16\right)-45=48-45=3$