Let \( a_1, a_2, a_3, \ldots \) be an A.P. If \( \dfrac{a_1+a_2+\ldots+a_{10}}{a_1+a_2+\ldots+a_p}=\dfrac{100}{p^2},\; p\neq 10 \), then \( \dfrac{a_{11}}{a_{10}} \) is equal to:
\( \dfrac{19}{21} \)
\( \dfrac{100}{121} \)
\( \dfrac{121}{100} \)
\( \dfrac{21}{19} \)