Let the A.P. have first term $a$ and common difference $d$.

Then the sum of first $n$ terms is $S_n=\frac{n}{2}[2a+(n-1)d].$

Given $\frac{a_1+a_2+\cdots+a_{10}}{a_1+a_2+\cdots+a_p}=\frac{100}{p^2}, \qquad p\ne 10.$ That is, $\frac{S_{10}}{S_p}=\frac{100}{p^2}.$

Now compute $S_{10}$ and $S_p$: $S_{10}=\frac{10}{2}[2a+9d]=5(2a+9d),$ $S_p=\frac{p}{2}[2a+(p-1)d].$

So $\frac{5(2a+9d)}{\frac{p}{2}[2a+(p-1)d]}=\frac{100}{p^2}.$

Simplify: $\frac{10(2a+9d)}{p[2a+(p-1)d]}=\frac{100}{p^2}.$

Cross-multiplying, $10p(2a+9d)=100[2a+(p-1)d].$

Divide by $10$: $p(2a+9d)=10[2a+(p-1)d].$

Expand: $2ap+9pd=20a+10pd-10d.$

Rearranging, $2a(p-10)+d(9p-10p+10)=0,$ $2a(p-10)+d(10-p)=0.$

Since $10-p=-(p-10)$, $2a(p-10)-d(p-10)=0,$ $(p-10)(2a-d)=0.$

Given $p\ne 10$, hence $2a-d=0 \implies d=2a.$

Now find $\dfrac{a_{11}}{a_{10}}$.

In an A.P., $a_n=a+(n-1)d.$

Therefore, $a_{10}=a+9d=a+18a=19a,$ $a_{11}=a+10d=a+20a=21a.$

Hence, $\frac{a_{11}}{a_{10}}=\frac{21a}{19a}=\frac{21}{19}.$

Checking options:

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

So the correct option is C.