1. Tardigrade
  2. Question
  3. Mathematics
  4. Let α be the minimum value of f(x)=5 x2-2 x+(26/5) and the graph of f(x) is symmetric about x =β. Also, S n =α+(α+β)+(α+2 β)+(α+3 β)+ ldots ldots ldots upto n terms. The value of α is equal to

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Q. Let $\alpha$ be the minimum value of $f(x)=5 x^2-2 x+\frac{26}{5}$ and the graph of $f(x)$ is symmetric about $x =\beta$. Also, $S _{ n }=\alpha+(\alpha+\beta)+(\alpha+2 \beta)+(\alpha+3 \beta)+\ldots \ldots \ldots$ upto $n$ terms.
The value of $\alpha$ is equal to

Sequences and Series

Solution:

$f(x)=5 x^2-2 x+\frac{26}{5}=5\left(x-\frac{1}{5}\right)^2+5 $
$\therefore f_{\min }\left(x=\frac{1}{5}\right)=5 \equiv \alpha $