Let mathrm{a}_{1}, mathrm{a}_{2}, mathrm{a}_{ | vedclass

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Question

$\frac{\frac{10}{2}\left(2 \mathrm{a}_{1}+9 \mathrm{~d}\right)}{\frac{\mathrm{p}}{2}\left(2 \mathrm{a}_{1}+(\mathrm{p}-1) \mathrm{d}\right)}=\frac{100

Vedclass · Accepted Answer

$\frac{21}{19}$

Vedclass · Answer

$\frac{\frac{10}{2}\left(2 \mathrm{a}_{1}+9 \mathrm{~d}ight)}{\frac{\mathrm{p}}{2}\left(2 \mathrm{a}_{1}+(\mathrm{p}-1) \mathrm{d}ight)}=\frac{100}{\mathrm{p}^{2}}$

$\left(2 \mathrm{a}_{1}+9 \mathrm{~d}ight) \mathrm{p}=10\left(2 \mathrm{a}_{1}+(\mathrm{p}-1) \mathrm{d}ight)$

$9 \mathrm{dp}=20 \mathrm{a}_{1}-2 \mathrm{pa}_{1}+10 \mathrm{~d}(\mathrm{p}-1)$

$9 \mathrm{p}=(20-2 \mathrm{p}) \frac{\mathrm{a}_{1}}{\mathrm{~d}}+10(\mathrm{p}-1)$

$\frac{\mathrm{a}_{1}}{\mathrm{~d}}=\frac{(10-\mathrm{p})}{2(10-\mathrm{p})}=\frac{1}{2}$

$	herefore \frac{\mathrm{a}_{11}}{\mathrm{a}_{10}}=\frac{\mathrm{a}_{1}+10 \mathrm{~d}}{\mathrm{a}_{1}+9 \mathrm{~d}}=\frac{\frac{1}{2}+10}{\frac{1}{2}+9}=\frac{21}{19}$

Let $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots$ 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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