IF `a_(1),a_(2),a_(3),"...."a_(10)` be in AP and `h_(1),h_(2),h_(3),"...."h_(10)` be in HP. If `a_(1)=h_(1)=2` and `a_(10)=h_(10)=3`, then find value of `a_(4)h_(7)`.

Explanation

` therefore a_(1),a_(2),a_(3),"...."a_(10)` are in AP.
If d be the common difference, then `d=(a_(10)-a_(1))/(9)=(3-2)/(9)=(1)/(9)`
`therefore a_(4)=a_(1)+3d=2+(3)/(9)=2+(1)/(3)=(7)/(3) "……..(i)"`
and given ` h_(1),h_(2),h_(3),"...."h_(10)` are in HP.
If D be common difference of corresponding AP.
Then, ` D=((1)/h_(10)-(1)/h_(1))/(9)=((1)/(3)-(1)/(2))/(9)=-(1)/(54)`
` therefore (1)/h_(7)=(1)/h_(1)+6D=(1)/(2)-(6)/(54)=(1)/(2)-(1)/(9)=(7)/(18) implies h_(7)=(18)/(7)`
Hence, ` a_(4)*h_(7)=(7)/(3)xx(18)/(7)=6`