If 32sin2α−1,  14   and   34−2sin2α arethe first three terms of an A.P for some α ,  then the sixth term of this A.P. is:

let 32sin2α=t, t∈19,9Hence, t3,14,81t terms are in arithmetic progression. If a,b,c are in arithmetic progression then 2b=a+cit implies that 28=t3+81t⇒t2−84t+243=0⇒t=3,81suppose t=3, so that 32sin2α=3⇒sin2α=12when t=3 the terms are 1,14,27,...First term is 1, common difference is 13a6=a+5d=1+513=66