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

`:.((p)/(2)[2a_(1)+(p-1)d])/((q)/(2)[2a_(1)+(q-1)d])=(p^(2))/(q^(2))`
`implies (2a_(1)+(p-1)d)/(2a_(1)+(q-1)d)=(p)/(q)implies (a_(1)+((p-1)/(2))d)/(a_(1)+((q-1)/(2))d)=(p)/(q)`
For `(a_(6))/(a_(21)),p=11" and " q=41 implies(a_(6))/(a_(21))=(11)/(41)`.