If 2 a 1 , 2 a 2 , 2 a 3 . . . . . 2 a r are in geometric progression, then a 1 a 2 a 3 a n + 1 a n + 2 a n + 3 a 2 n + 1 a 2 n + 2 a 2 n + 3 is equal to

Step-by-step solution

2 a 1 , 2 a 2 , 2 a 3 . . . . . are in G . P ⇒ a 2 - a 1 = a 3 - a 2 = a 4 - a 3 = . . . . . ⇒ a 1 , a 2 , a 3 . . . . . are in A . P . ⇒ a 2 n + 1 + a 1 = 2 a n + 1 , a 2 + a 2 n + 2 = 2 a n + 2 , a 3 + a 2 n + 3 = 2 a n + 3 So, by operation R 2 → R 2 - 1 2 ( R 1 + R 3 ) ⇒ Δ = a 1 a 2 a 3 0 0 0 a 2 n + 1 a 2 n + 2 a 2 n + 3 = 0