# Let a1, a2, a3,.......... be a A. P. if

Question:

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 :

1. $\frac{19}{21}$

2. $\frac{100}{121}$

3. $\frac{21}{19}$

4. $\frac{121}{100}$

Correct Option: 3

Solution:

$\frac{\frac{10}{2}\left(2 \mathrm{a}_{1}+9 \mathrm{~d}\right)}{\frac{\mathrm{p}}{2}\left(2 \mathrm{a}_{1}+(\mathrm{p}-1) \mathrm{d}\right)}=\frac{100}{\mathrm{p}^{2}}$

$\left(2 \mathrm{a}_{1}+9 \mathrm{~d}\right) \mathrm{p}=10\left(2 \mathrm{a}_{1}+(\mathrm{p}-1) \mathrm{d}\right)$

$9 \mathrm{dp}=20 \mathrm{a}_{1}-2 \mathrm{pa}_{1}+10 \mathrm{~d}(\mathrm{p}-1)$

$9 p=(20-2 p) \frac{a_{1}}{d}+10(p-1)$

$\frac{a_{1}}{d}=\frac{(10-p)}{2(10-p)}=\frac{1}{2}$

$\therefore \frac{\mathrm{a}_{11}}{\mathrm{a}_{10}}=\frac{\mathrm{a}_{1}+10 \mathrm{~d}}{\mathrm{a}_{1}+9 \mathrm{~d}}=\frac{\frac{1}{2}+10}{\frac{1}{2}+9}=\frac{21}{19}$