Show that the twenty-first,thirty-seventh and sixty-fifth term of a linear sequnce are consecutive terms of an exponential sequnce whose common ratio is 7/4

Answer 1

(a+36d)/(a+20d) = 7/4
(a+64d)/(a+36d) = 7/4

3a=4d

looks like there are many such sequences:

a=12, d=9
T21 = 192
T37 = 336
T65 = 588