Let `A_r ,r=1,2,3, ` , be the points on the number line such that `O A_1,O A_2,O A_3dot` are in `G P ,` where `O` is the origin, and the common ratio of the `G P` be a positive proper fraction. Let `M ,` be the middle point of the line segment `A_r A_(r+1.)` Then the value of `sum_(r=1)^ooO M_r` is equal to
(a) `(O A_1(O S A_1-O A_2))/(2(O A_1+O A_2))` (b) `(O A_1(O A_2+O A_1))/(2(O A_1-O A_2)` (c)`(O A_1)/(2(O A_1-O A_2))` (d) `oo`

Answer

B

Explanation

`OM_r=OA_r+(OA_(r+1)-OA_(r))/(2)`
`=(OA_(r)+OA_(r+1))/(2)`
`=(1)/(2){OA_1xxk^(r-1)+OA_(1)xxK^(r)}`
`=(OA_1)/(2)(1+k)k^(r-1)`
`therefore sum_(r=1)^(oo)OM_(r)=(OA)/(2)(1+k)sum_(r=1)^(oo)k^(r-1)`
`=(OA_1))/(2)(1+K)xx(1)/(1-k)`
`(OA)/(2)xx(1+(OA_2//OA_1))/(1-(OA_2//OA_1))=(OA _(1)(OA_1+OA_2))/(2(OA_1-OA_2))`