JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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\frac{1 \times 2^2+2 \times 3^2+3 \times 4^2+----+11(12)^2}{1^2 \times 2+2^2 \times 3+3^2 \times 4+----+11^2(12)}=

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Correct option: A

\frac{19}{17}

Option D (

3 x^2-20 x+16=0

) is correct because after simplifying the given equation and finding the new roots, the resulting quadratic equation is

3x^2 - 20x + 16 = 0

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\log \left(\frac{a}{3 b}\right)

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\frac{3}{4}