Let $${a_1},{a_2},......{a_{10}}$$    be in A.P. and $${h_1},{h_2},......{h_{10}}$$    be in H.P. If $${a_1} = {h_1} = 2$$   and $${a_{10}} = {h_{10}} = 3,$$   then $${a_4}{h_7}$$  is

Solution :
$$\eqalign{ & {a_1} = {h_1} = 2,{a_{10}} = {h_{10}} = 3 \cr & 3 = {a_{10}} = 2 + 9d \cr & \Rightarrow \,d = \frac{1}{9} \cr & \therefore \,\,{a_4} = 2 + 3d = \frac{7}{3} \cr & \,\,\,\,\,\,\,3 = {h_{10}} \Rightarrow \frac{1}{3} = \frac{1}{{{h_{10}}}} = \frac{1}{2} + 9D \cr & \therefore \,\,D = - \frac{1}{{54}} \cr & \,\,\,\,\,\frac{1}{{{h_7}}} = \frac{1}{2} + 6D = \frac{1}{2} - \frac{1}{9} = \frac{7}{{18}} \cr & \therefore \,\,{a_4}{h_7} = \frac{7}{3} \times \frac{{18}}{7} = 6. \cr} $$