MathematicsMedium178×since 2002Q5222Let a₁, a₂, a₃, ..... be an A.P. If a1+a2+....+a10a1+a2+....+ap=100p2{{{a_1} + {a_2} + .... + {a_{10}}} \over {{a_1} + {a_2} + .... + {a_p}}} = {{100} \over {{p^2}}}a1+a2+....+apa1+a2+....+a10=p2100, p ≠\ne= 10, then a11a10{{{a_{11}}} \over {{a_{10}}}}a10a11 is equal to :A1921{{19} \over {21}}2119B100121{{100} \over {121}}121100C2119{{21} \over {19}}1921D121100{{121} \over {100}}100121Check answerSkip