Let a 1 ,   a 2 ,   a 3 ,   … be an A.P. If a 1 + a 2 + … + a 10 a 1 + a 2 +…

Given, $\frac{a_1+a_2+\dots +a_{10}}{a_1+a_2+\dots +a_p}=\frac{S_{10}}{S_p}=\frac{100}{p^2}$

$\Rightarrow \frac{{\displaystyle \frac{10}{2}}\left[2a_1+9d\right]}{{\displaystyle \frac{p}{2}}\left[2a_1+\left(p-1\right)d\right]}=\frac{100}{p^2}$

$\Rightarrow \frac{{\displaystyle 10\left[2a_1+9d\right]}}{{\displaystyle p\left[2a_1+\left(p-1\right)d\right]}}=\frac{100}{p^2}$

$\Rightarrow \frac{{\displaystyle 2a_1+9d}}{{\displaystyle 2a_1+\left(p-1\right)d}}=\frac{10}{p}$

Let, $a_1=a$

$\Rightarrow 2pa+9pd=20a+10pd-10d$

$\Rightarrow 2pa-20a=pd-10d$        $$

$\Rightarrow 2a\left(p-10\right)=d\left(p-10\right)$

$\Rightarrow 2a=d$

$\Rightarrow \frac{a}{d}=\frac{1}{2}$

Now, $\frac{a_{11}}{a_{10}}=\frac{a+10d}{a+9d}$

$\Rightarrow \frac{a_{11}}{a_{10}}=\frac{d\left({\displaystyle \frac{a}{d}}+10\right)}{d\left({\displaystyle \frac{a}{d}}+9\right)}$

$\Rightarrow \frac{a_{11}}{a_{10}}=\frac{\frac{1}{2}+10}{\frac{1}{2}+9}$

$\Rightarrow \frac{a_{11}}{a_{10}}=\frac{21}{19}$