Question

Let $a_1,a_2,a_3,\dots$ be an A.P. If $\frac{a_1+a_2+\dots +a_{10}}{a_1+a_2+\dots +a_p}=\frac{100}{p^2},p\ne 10$, 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)

$\frac{\frac{10}{2}\left(2a_1+9d\right)}{\frac{p}{2}\left(2a_1+(p-1)d\right)}=\frac{100}{p^2}$
$\left(2a_1+9d\right)p=10\left(2a_1+(p-1)d\right)$
$9dp=20a_1-2p{a}_1+10d(p-1)$
$9p=(20-2p)\frac{a_1}{d}+10(p-1)$
$\frac{a_1}{d}=\frac{(10-p)}{2(10-p)}=\frac{1}{2}$
$\therefore \frac{a_{11}}{a_{10}}=\frac{a_1+10d}{a_1+9d}$
$=\frac{\frac{1}{2}+10}{\frac{1}{2}+9}=\frac{21}{19}$