1.

let `a_1,a_2,a_3,...........,`be an `AP` such that `(a_1+a_2+a_3+...........+a_p)/(a_1+a_2+a_3+...........+a_q)=p^3/q^3`,`(p!=q)` then find `a_6/a_21=?`A. `(41)/(11)`B. `(7)/(2)`C. `(2)/(7)`D. `(11)/(41)`

Answer» Correct Answer - D
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)`.


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