Question

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_4h_7` is
A. 2
B. 3
C. 5
D. 6

Answer 1

Correct Answer - D
Let d be the common difference of the A.P. Then,
`a_(10)=3rArra_(1)+9d=3rArr2+9d=3rArrd=(1)/(9)`.
`:." "a_(4)=a_(1)+3d=2+(1)/(3)=(7)/(3)`
Let D be the common difference of the A.P. `(1)/(h_(1)),(1)/(h_(2)), . . . .,(1)/(h_(10))`.
Then,
`h_(10)=3`
`rArr" "(1)/(h_(10))=(1)/(3)`
`rArr" "(1)/(h_(1))+9D=(1)/(3)rArr(1)/(2)+9D=(1)/(3)rArr9D=-(1)/(6)rArrD=-(1)/(54)`
`:." "(1)/(h_(7))=(1)/(h_(1))+6D=(1)/(2)-(1)/(9)=(7)/(18)rArrh_(7)=(18)/(7)`
Hence, `a_(4)h_(7)=(7)/(3)xx(18)/(7)=6`.