Let $\alpha, \beta (\alpha \neq \beta)$ be the values of $m$, for which the equations $\mathrm{x}+\mathrm{y}+\mathrm{z}=1 ; \mathrm{x}+2 \mathrm{y}+4 \mathrm{z}=\mathrm{m}$ 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 :

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Determinants

Let

α, β (α \neq = β)

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} (n^{α} + n^{β})

is equal to :

Answer: (A)

For infinitely many solutions the coefficient determinant must vanish:

1111241410 = 1 (20 - 16) - 1 (10 - 4) + 1 (4 - 2) = 0

Indeed row 3 is a combination of rows 1 and 2: solving

a + b = 1, a + 2 b = 4

gives

b = 3, a = - 2

, and checking the third entry

a + 4 b = - 2 + 12 = 10

confirms row

_{3} = - 2 \cdot

row

_{1} + 3 \cdot

row

_{2}

.

For consistency, the right-hand sides must obey the same relation:

m^{2} = - 2 (1) + 3 m \Rightarrow m^{2} - 3 m + 2 = 0 \Rightarrow (m - 1) (m - 2) = 0

{α, β} = {1, 2}

, and

n = 1 \sum 10 (n^{α} + n^{β}) = n = 1 \sum 10 n + n = 1 \sum 10 n^{2} = 55 + 385 = 440

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Single correct option