If a 1 , a 2 , a 3 , a 4 , a 5 are consecutive terms of an arithmetic progression with common difference 3, then the value of a 3 2 a 2 a 1 a 4 2 a 3 a 2 a 5 2 a 4 a 3 is

Step-by-step solution

Apply R 3 ↔ R 3 - R 2 ,   R 2 ↔ R 2 - R 1 a 3 2 a 2 a 1 a 4 2 - a 3 2 a 3 - a 2 a 2 - a 1 a 5 2 - a 4 2 a 4 - a 3 a 3 - a 2 = a 3 2 a 2 a 1 3 a 3 + a 4 3 3 3 a 4 + a 5 3 3 Applying C 3 ↔ C 3 - C 2 a 3 2 a 2 - 3 3 a 3 + a 4 3 0 3 a 4 + a 5 3 0 = - 3 9 a 3 + a 4 - 9 a 4 + a 5 = - 27 a 3 - a 5 = - 27 × - 6 = 162