Question

Let $a_{1}, a_{2}, a_{3}, \ldots$ be an A.P. If $\frac{a_{1}+a_{2}+\ldots+a_{10}}{a_{1}+a_{2}+\ldots+a_{p}}=\frac{100}{p^{2}}, p \neq 10$, then $\frac{a_{11}}{a_{10}}$ is equal to:

• A. $\frac{19}{21}$
• B. $\frac{100}{121}$
• C. $\frac{21}{19}$
• D. $\frac{121}{100}$

Answer 1

Answer: (C)

$\frac{\frac{10}{2}\left(2 a_{1}+9 d\right)}{\frac{p}{2}\left(2 a_{1}+(p-1) d\right)}=\frac{100}{p^{2}}$
$\left(2 a_{1}+9 d\right) p=10\left(2 a_{1}+(p-1) d\right)$
$9 d p=20 a_{1}-2 p a_{1}+10 d(p-1)$
$9 p=(20-2 p) \frac{a_{1}}{d}+10(p-1)$
$\frac{a_{1}}{d}=\frac{(10-p)}{2(10-p)}=\frac{1}{2}$
$\therefore \frac{a_{11}}{a_{10}}=\frac{a_{1}+10 d}{a_{1}+9 d}$
$=\frac{\frac{1}{2}+10}{\frac{1}{2}+9}=\frac{21}{19}$