Question

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

The correct option is C $\frac{21}{19}$

$\because a_1,a_2,a_3,....$ are in A.P.
Let its common difference be $d$.
$\therefore \frac{a_1+a_2+....+a_{10}}{a_1+a_2+....+a_p}=\frac{100}{p^2}$
$\Rightarrow \frac{\frac{10}{2}\text{{ 2a}{}_1\text{+ 9d}}}{\frac{p}{2}\left\{2a_1+(p-1)d\right\}}=\frac{100}{p^2}$
$\Rightarrow \frac{2a_1+9d}{2a_1+(p-1)d}=\frac{10}{p}$
$\Rightarrow 2p{a}_1+9pd=20a_1+10(p-1)d$
$\Rightarrow (2p-20)a_1=(p-10)d$
$\Rightarrow 2a_1(p-10)-d(p-10)=0$
$\therefore 2a_1=d\because p\ne 10$
$\therefore \frac{a_{11}}{a_{10}}=\frac{a_1+10d}{a_1+9d}=\frac{a_1+20a_1}{a_1+18a_1}=\frac{21}{19}$