Let a_{1}, a_{2}, a_{3}, ldots be an A.P. If | vedclass

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(C) The sum of the first $n$ terms of an $A.P.$ is given by $S_{n} = \frac{n}{2}(2a_{1} + (n-1)d)$.Given $\frac{S_{10}}{S_{p}} = \frac{100}{p^{2}}$,we

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$\frac{21}{19}$

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(C) The sum of the first $n$ terms of an $A.P.$ is given by $S_{n} = \frac{n}{2}(2a_{1} + (n-1)d)$.Given $\frac{S_{10}}{S_{p}} = \frac{100}{p^{2}}$,we have $\frac{\frac{10}{2}(2a_{1} + 9d)}{\frac{p}{2}(2a_{1} + (p-1)d)} = \frac{100}{p^{2}}$.Simplifying,$\frac{5(2a_{1} + 9d)}{\frac{p}{2}(2a_{1} + (p-1)d)} = \frac{100}{p^{2}} \implies \frac{2a_{1} + 9d}{2a_{1} + (p-1)d} = \frac{10}{p}$.Cross-multiplying gives $p(2a_{1} + 9d) = 10(2a_{1} + (p-1)d)$.$2a_{1}p + 9dp = 20a_{1} + 10dp - 10d$.$10d - 20a_{1} = 10dp - 9dp - 2a_{1}p = dp - 2a_{1}p = p(d - 2a_{1})$.Alternatively,$2a_{1}(p - 10) = d(10p - 9p - 10) = d(p - 10)$.Since $p 
eq 10$,we divide by $(p - 10)$ to get $2a_{1} = d$,or $\frac{a_{1}}{d} = \frac{1}{2}$.We need to find $\frac{a_{11}}{a_{10}} = \frac{a_{1} + 10d}{a_{1} + 9d}$.Substituting $d = 2a_{1}$,we get $\frac{a_{1} + 10(2a_{1})}{a_{1} + 9(2a_{1})} = \frac{a_{1} + 20a_{1}}{a_{1} + 18a_{1}} = \frac{21a_{1}}{19a_{1}} = \frac{21}{19}$.

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 :

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