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JEE Main 202529 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Determinants

JEE Main 2025 · 29 January · Evening Shift · Mathematics · Determinants

Let α, β(α ≠ β) be the values of m , for which the equations...

Let α,β(α≠β)\alpha, \beta(\alpha \neq \beta) be the values of m , for which the equations x+y+z=1;x+2y+4z=mandx+y+z=1 ; x+2 y+4 z=\mathrm{m} andx+4 y+10 z=m^2haveinfinitelymanysolutions.Thenthevalueofhave infinitely many solutions. Then the value of\sum_{n=1}^{10}(nα+nβ)\left(n^\alpha+n^\beta\right) is equal to :

Options

  1. A3080
  2. B560
  3. C3410
  4. D440

Correct answer

D. 440

Step-by-step solution

Δ=∣1111241410∣=1(20−16)−1(10−4)+1(4−2)=4−6+2=0\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 Δx=Δy=Δz=0m2−3x+2=0m=1,2α=1,β=2∴∑n=110(nα+nβ)=∑n=110n1+∑n=110n2=10(11)2+10(11)(21)6=55+385=440\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}
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