Let α, β(α ≠ β) be values of m , for which equations x+y+z=1 ; x+2 y+4 z=m and x+4 y+10 z=m^2

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3 and 4 .Determinants and Matrices

hard

Let $\alpha, \beta(\alpha \neq \beta)$ be the values of $m$, for which the equations $x+y+z=1 ; x+2 y+4 z=m$ and $x+4 y+10 z=m^2$ have infinitely many solutions. Then the value of $\sum_{n=1}^{10}\left(n^\alpha+n^\beta\right)$ is equal to :

A$560$

B$3080$

C$3410$

D$440$

(JEE MAIN-2025)

Solution

$\Delta=\left|\begin{array}{llc}1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10\end{array}\right|=1(20-16)-1(10-4)+1(4-2)$
$=4-6+2=0$
For infinite solutions
$\Delta_{x}=\Delta_{y}=\Delta_z=0$
$m^2-3 x+2=0$
$m=1,2$
$\alpha=1, \beta=2$
$\therefore \sum_{n=1}^{10}\left(n^\alpha+n^\beta\right)=\sum_{n=1}^{10} n^1+\sum_{n=1}^{10} n^2$
$= \frac{10(11)}{2}+\frac{10(11)(21)}{6}$
$= 55+385$
$= 440$

Standard 12

Mathematics

A root of the equation $\left| {\,\begin{array}{*{20}{c}}{3 – x}&{ – 6}&3\\{ – 6}&{3 – x}&3\\3&3&{ – 6 – x}\end{array}\,} \right| = 0$ is

easy

In a third order determinant, each element of the first column consists of sum of two terms, each element of the second column consists of sum of three terms and each element of the third column consists of sum of four terms. Then it can be decomposed into $n $determinants, where $ n$ has the value

medium

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normal

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medium

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medium

(IIT-1983)

Let $\alpha, \beta(\alpha \neq \beta)$ be the values of $m$, for which the equations $x+y+z=1 ; x+2 y+4 z=m$ and $x+4 y+10 z=m^2$ have infinitely many solutions. Then the value of $\sum_{n=1}^{10}\left(n^\alpha+n^\beta\right)$ is equal to :

Solution

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