Let a be the first term and r be the common ratio, then according to question $a_4$ = $ar^3$ = x ...(i) $a_{10}$ = $ar^9$ = y ...(ii) $a_{16}$ = $ar^{15}$ = z ...(iii) Multiplying (i) and (iii), we get $a^2 r^{18}$ = xz ⇒ $(ar^9)^2$ = xz ⇒ $y^2$ = xz [using (ii)] Hence x, y, z are in G.P.