If `a_1, a_2, a_3, ` be terms of an A.P. and `(a_1+a_2+.....+a_p)/(a_1+a_2+.....+a_q)=(p^2)/(q^2), p!=q ,t h e n(a_6)/(a_(21))` equals to (a).41/11`` (b). 7/2`` (c). 2/7`` (d). 11/41

Options

A. `(41)/(11)`
B. `(7)/(2)`
C. `(2)/(7)`
D. `(11)/(41)`

Answer

D

Explanation

We have,
`(a_(1)+a_(2)+ . . .+a_(p))/(a_(1)+a_(2)+ . . ..+a_(q))=(p^(2))/(q^(2))rArr(a_(1)+((p-1)/(2))d)/(a_(1)+((q-1)/(2))d)=(p)/(q)` . . . .(i) We have to find the value of `(a_(6))/(a_(21))` i.e. `(a_(1)+5d)/(a_(1)+20)`.
So, replacing `(p-1)/(2)` by 5 and `(q-1)/(2)=20` i.e. putting p=11 and q=41 in (i), we get `(a_(6))/(a_(12))=(11)/(41)`.