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JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Determinants

For

α, β \in R

and a natural number

n

, let

A_{r} = r 2 r 3 r - 2 123 \frac{n ^{2}}{2} + α n^{2} - β \frac{n ( 3 n - 1 )}{2}

. Then

2 A_{10} - A_{8}

is

(A) $2 α + 4 β$
(B) $2 n$
(C)0
(D) $4 α + 2 β$ ✓ Correct

Answer: (D)

Solution

Columns 2 and 3 of

A_{r}

do not depend on

r

, so

2 A_{10} - A_{8}

is the determinant with first column

2102028 - 81622 = 122434

.
Expanding with third-column entries

a = \frac{n ^{2}}{2} + α, b = n^{2} - β, c = \frac{n ( 3 n - 1 )}{2}

:

det = 12 (2 c - 3 b) - (24 c - 34 b) + a (72 - 68) = - 2 b + 4 a = 4 α + 2 β .

JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Single correct option

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