Let the numbers 2 , b , c be in an A.P. and A = 1 1 1 2 b c 4 b 2 c 2 . If d e t ( A ) ∈ [ 2,16 ] , then c lies in the interval:

Step-by-step solution

1 1 1 2 b c 4 b 2 c 2 → C 2 → C 2 - C 1 ,   C 3 → C 3 - C 1 1 0 0 2 b - 2 c - 2 4 b 2 - 4 c 2 - 4 = b - 2 c - 2 c - b Let common difference of A P = d ∴   b = 2 + d ,       c = 2 + 2 d A = d × 2 d × d = 2 d 3 ∈ 2 ,   16 ⇒ d 3 ∈ 1 ,   8 ⇒ d ∈ 1 ,   2 ⇒ 2 d ∈ 2 ,   4 ⇒ 2 + 2 d ∈ 4 ,   6 ∴   c ∈ 4 ,   6