Skip to main content
Back to SEQUENCE OF SERIES

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy
1 of 1
Analyzing +4 | -1
If x=log⁡2a(bcd2),y=log⁡3b(acd3),x=\log _{2 a}\left(\frac{b c d}{2}\right), y=\log _{3 b}\left(\frac{a c d}{3}\right), z=log⁡4c(abd4)z=\log _{4 c}\left(\frac{a b d}{4}\right) and w=log⁡5d(abc5)w=\log _{5 d}\left(\frac{a b c}{5}\right) and 1x+1+1y+1+1z+1+1w+1=log⁡abcdN+1,\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}+\frac{1}{w+1}=\log _{a b c d} N+1, the value of NN is
View answer and explanation
Correct option: C
120
Option C (120) is correct because by simplifying the given logarithmic expressions and summing them, we find that N = 120.

More Questions from SEQUENCE OF SERIES

Practice All Chapter MCQs →
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.