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=2a n da_(10)=h_(10)=3,t h e na_4h_7` is `2` b. `3` c. `5` d. `6`

Options

A. 2
B. 3
C. 5
D. 6

Answer

D

Explanation

Given that `a_(1),a_(2),"……"a_(10)` be un AP.
Let d be the common difference of AP.
`:. d=(a_(10)-a_(1))/(10-1)`
`d=(3-2)/(9) " " [" given that " a_(1)=h_(1)=2 " and " a_(10)=h_(10)=3]`
`d=(1)/(9)`
`:.a_(4)=a_(1)+3d=2+(3)/(9)=2+(1)/(3)=(7)/(3)`
Now, `h_(1),h_(2),"......,"h_(10)` be in HP.
So, common difference of respective AP.
`D=((1)/(h_(10))-(1)/(h_(1)))/(10-1)=((1)/(3)-(1)/(2))/(9)=(-1)/(9xx6)-(-1)/(54)`
So, `(1)/(h_(7))=(1)/(h_(1))+6D " " implies (1)/(lambda_(7))=(1)/(2)+6((-1)/(54))=(1)/(2)-(1)/(9)`
`(1)/(h_(7))=(7)/(18)" " implies h_(7)=(18)/(7)`
So, `a_(4)h_(7)=(7)/(3)xx(18)/(7)=6`.