Let the tine number is in AP are (a−2d),(a−d),a,(a+d),(a+2d)
where, d=0
Given, 1 st, 3rd and 4th terms are in GP. ⇒a2=(a−2d)(a+d) ⇒a2=a2−2ad+ad−2d2 ⇒2d2+ad=0 ⇒d(2d+a)=0 ∵d=0 ∴a+2d=0 ⇒a=−2d
Hence, terms are −4d,−3d,−2d,−d,0 ∴ The fifth term is always 0 .