Q43 / 75 · JEE Main 2025 (29 Jan Shift 2) · Mathematics

Let $\alpha, \beta(\alpha \neq \beta)$ be the values of m , for which the equations $x+y+z=1 ; x+2 y+4 z=\mathrm{m}$ 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
    3080
  • B
    560
  • C
    3410
  • Correct answer
    D
    440

Solution

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

Hence, the correct answer is option: 'D'.

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