Let be the values of m, for which the equations x + y + z = 1; x + 2y +4z = m and x + 4y + 10z = have infinitely many solutions. Then the value of is equal to : [2025]

Question

Let $α, β (α \neq β)$ be the values of m, for which the equations x + y + z = 1; x + 2y +4z = m and x + 4y + 10z = $m^{2}$ have infinitely many solutions. Then the value of $\sum_{n = 1}^{10} (n^{α} + n^{β})$ is equal to : [2025]

Smart Achievers · Accepted Answer

(4)

We have, x + y + z = 1

x + 2y +4z = m and x + 4y + 10z = $m^{2}$

$∴ △ =$ $| \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10 \end{matrix} |$ $= 1 (4) - 1 (6) + 1 (2) = 0$

For infinite many solutions, $△ x = △ y = △ z = 0$

Now, $△ x = 0 \Rightarrow$ $| \begin{matrix} 1 & 1 & 1 \\ m & 2 & 4 \\ m^{2} & 4 & 10 \end{matrix} |$ $= 0$

$\Rightarrow 1 (4) - 1 (10 m - 4 m^{2}) + 1 (4 m - 2 m^{2}) = 0$

$\Rightarrow 4 - 10 m + 4 m^{2} + 4 m - 2 m^{2} = 0$

$\Rightarrow 2 m^{2} - 6 m + 4 = 0$

$\Rightarrow m^{2} - 3 m + 2 = 0 \Rightarrow m = 1, 2$

$∴ α = 1, β = 2$

Now, $\sum_{n = 1}^{10} (n^{α} + n^{β}) = \sum_{n = 1}^{10} n + \sum_{n = 1}^{10} n^{2}$

$= \frac{10 (11)}{2} + \frac{10 (11) (21)}{6}$ = 55 + 385 = 440.

Solution

Q.
Let $α, β (α \neq β)$ be the values of m, for which the equations x + y + z = 1; x + 2y +4z = m and x + 4y + 10z = $m^{2}$ have infinitely many solutions. Then the value of $\sum_{n = 1}^{10} (n^{α} + n^{β})$ is equal to : [2025]

1 3080

2 560

3 3410

4 440

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