If first fourth and nine term ofAP are in GP find the numbers
for AP
first -- a
fourth --a+3d
ninth -- a+8d
these are to be in GP
(a+3d)^2 = a(a+8d)
a^2 + 6ad + 9d^2 = a^2 + 8ad
9d^2 - 2ad = 0
d(9d -2a) = 0
d = 0 or a = 9d/2
reject d=0, because the sequence simply becomes
a, a, a, .... which technically is an AP as well as a GP
if a = 9d/2, we can pick any value of d, d≠0, and we will get a solution
e.g. let d = 2, then a = 9
and our sequence is
9, 11, 13, 15, 17, 19, 21, 23, 25, ...
is 15/9 = 25/15 ?
is 5/3 = 5/3 , yes!
let d = 6, a = 27
our sequence is
27, 33, 39, 45, ....
term9 = 27+8(6) = 75
check: term9/term4 = 75/45 = 5/3
term4/term1 = 45/27 = 5/3 , yes!