If 3 2 sin 2 α - 1 , 14 and 3 4 - 2 sin 2 α are the first three terms of an A.P. for some α , then the sixth term of this A.P. is

Step-by-step solution

If a ,   b ,   c are in AP, then b is A.M of a   &   c . ∴   2   b = a + c ⇒ 28 = 3 2 sin 2 α - 1 + 3 4 - 2 sin 2 α Putting, 3 2 sin 2 α = x we get, 28 = x 3 + 81 x ⇒ x 2 - 84 x + 243 = 0 ⇒ ( x - 3 ) ( x - 81 ) = 0 ∴   3 2 sin 2 α = 3 or 3 4 sin 2 α = 1 2 , ∵ sin 2 α ≠ 2 Terms are 1 , 14 , 27 , … … then T 6 = 1 + 5 ( 13 ) = 66