Let α, β ( α neq β ) be the values of m for which the equations x + y + z = 1, x + 2y + 4z = m, x + 4y + 10z = m 2 have infinitely many solutions. Then the value of sum n=1 10 ( α n + β n ) is equal to:

Mathematics · Determinants

JEE Main 2025 — 29 January, Evening Shift — Question 52

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

Option A:
440
Correct
Option B:
3080
Option C:
3410
Option D:
560

Answer: A

Step-by-step solution

\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10 \end{vmatrix} = 1(20 - 16) - 1(10 - 4) + 1(4 - 2)

= 4 - 6 + 2 = 0

For infinitely many solutions,

\Delta_x = \Delta_y = \Delta_z = 0

m^2 - 3m + 2 = 0

m = 1,\ 2

\alpha = 1,\quad \beta = 2

\therefore \sum_{n=1}^{10} (\alpha^n + \beta^n) = \sum_{n=1}^{10} 1^n + \sum_{n=1}^{10} 2^n

= \frac{10(11)}{2} + \frac{10(11)(21)}{6}

= 55 + 385

= 440

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 29 January, Evening Shift
Subject: Mathematics
Chapter: Determinants
Topic: Consistency of Non-homogeneous system