JEE Mains · Maths · STD 11 - 8. sequence and series
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 :
- A \(\frac{19}{21}\)
- B \(\frac{100}{121}\)
- C \(\frac{21}{19}\)
- D \(\frac{121}{100}\)
Answer & Solution
Correct Answer
(C) \(\frac{21}{19}\)
Step-by-step Solution
Detailed explanation
\(\frac{\frac{10}{2}\left(2 \mathrm{a}_{1}+9 \mathrm{~d}\right)}{\frac{\mathrm{p}}{2}\left(2 \mathrm{a}_{1}+(\mathrm{p}-1) \mathrm{d}\right)}=\frac{100}{\mathrm{p}^{2}}\)…
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