Let $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots$ be an A.P. If
$\frac{\mathrm{a}_{1}+\mathrm{a}_{2}+\ldots+\mathrm{a}_{10}}{\mathrm{a}_{1}+\mathrm{a}_{2}+\ldots+\mathrm{a}_{\mathrm{p}}}=\frac{100}{\mathrm{p}^{2}}, \mathrm{p} \neq 10$, then $\frac{\mathrm{a}_{11}}{\mathrm{a}_{10}}$ is equal
to :
Correct Option: 3
$\frac{\frac{10}{2}\left(2 \mathrm{a}_{1}+9 \mathrm{~d}\right)}{\frac{\mathrm{p}}{2}\left(2 \mathrm{a}_{1}+(\mathrm{p}-1) \mathrm{d}\right)}=\frac{100}{\mathrm{p}^{2}}$
$\left(2 \mathrm{a}_{1}+9 \mathrm{~d}\right) \mathrm{p}=10\left(2 \mathrm{a}_{1}+(\mathrm{p}-1) \mathrm{d}\right)$
$9 \mathrm{dp}=20 \mathrm{a}_{1}-2 \mathrm{pa}_{1}+10 \mathrm{~d}(\mathrm{p}-1)$
$9 p=(20-2 p) \frac{a_{1}}{d}+10(p-1)$
$\frac{a_{1}}{d}=\frac{(10-p)}{2(10-p)}=\frac{1}{2}$
$\therefore \frac{\mathrm{a}_{11}}{\mathrm{a}_{10}}=\frac{\mathrm{a}_{1}+10 \mathrm{~d}}{\mathrm{a}_{1}+9 \mathrm{~d}}=\frac{\frac{1}{2}+10}{\frac{1}{2}+9}=\frac{21}{19}$