8. Sequences and Series
hard

અહી $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots$ એ સમાંતર શ્રેણીમાં છે. જો  $\frac{a_{1}+a_{2}+\ldots+a_{10}}{a_{1}+a_{2}+\ldots+a_{p}}=\frac{100}{p^{2}}, p \neq 10$ હોય તો  $\frac{a_{11}}{a_{10}}$ ની કિમંત મેળવો.

A

$\frac{19}{21}$

B

$\frac{100}{121}$

C

$\frac{21}{19}$

D

$\frac{121}{100}$

(JEE MAIN-2021)

Solution

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

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

$9 \mathrm{dp}=20 \mathrm{a}_{1}-2 \mathrm{pa}_{1}+10 \mathrm{~d}(\mathrm{p}-1)$

$9 \mathrm{p}=(20-2 \mathrm{p}) \frac{\mathrm{a}_{1}}{\mathrm{~d}}+10(\mathrm{p}-1)$

$\frac{\mathrm{a}_{1}}{\mathrm{~d}}=\frac{(10-\mathrm{p})}{2(10-\mathrm{p})}=\frac{1}{2}$

$\therefore \frac{\mathrm{a}_{11}}{\mathrm{a}_{10}}=\frac{\mathrm{a}_{1}+10 \mathrm{~d}}{\mathrm{a}_{1}+9 \mathrm{~d}}=\frac{\frac{1}{2}+10}{\frac{1}{2}+9}=\frac{21}{19}$

Standard 11
Mathematics

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