The numbers 32sin⁡2α−1,14 and 34−2sin⁡2α from first three terms of an A.P its 5th  term is

28=32sin⁡2α−1+34−2sin⁡2α  since 2b=a+c  Let a=32sin⁡2α28=a3+34a→a2−84a+243=0→a=81,3  If a=3→32sin⁡2α=31 if a=81 then sin2α=2 which is not possible →sin⁡2α=1/2→2α=30α=15∘ the first three terms are 1,14,27 Then t5=a+4d=1+52=53