In a geometric sequence, whose first term is #a_1# and common ratio is #r#, while #n^(th)# term #a_n=a_ar^((n-1))#, sum of first #n# terms is given by #S_n=(a_1(r^n-1))/(r-1)#.
Here assume that first term is #a_1# and common ratio is #r#. As sum of first four terms is #15#
#S_4=(a_1(r^4-1))/(r-1)=15# .........................(A)
i.e. #a_1+a_1r+a_1r^2+a_1r^3=15#
i.e. #a_1(1+r+r^2+r^3)=15# .........................(1)
and #a_1(r^4+r^5+r^6+r^7)=240#
i.e. #a_1r^4(1+r+r^2+r^3)=240# .........................(2)
Dividing (2) be (1), we get
#r^4=240/15=16#
i.e. #r=+-2#
If #r=2#, #a_1=15/(1+2+4+8)=1#
and if #r=-2#, #a_1=15/(1-2+4-8)=15/(-5)=-3#.
Hence sequence is
either #{1,2,4,8,16,32,64,18}# i.e. #a_1=1# and #r=2#
or #{-3,+6,-12,+24,-48,96,-192,384}# i.e. #a_1=-3# and #r=-2#
