MHT CET · Maths · Sequences and Series
If \(\frac{2+4+6+8-------\text { upto } n \text { terms }}{1+3+5+7-------\text { upto } n \text { terms }}=\frac{37}{36}\), then \(n=\)
- A 36
- B 29
- C 23
- D 37
Answer & Solution
Correct Answer
(A) 36
Step-by-step Solution
Detailed explanation
sum of \(n\) even natural numbers \(=n(n+1)\)
sum of \(n\) odd natural numbers \(=n^{\wedge} 2\)
\(n(n+1) / n^{\wedge} 2=37 / 36\)
\(n+1 / n=37 / 36\)
cross multiplication
\(36(n+1)=37 n\)
\(36 n+36=37 n\)
\(37 n-36 n=36\)
\(n=36\)
sum of \(n\) odd natural numbers \(=n^{\wedge} 2\)
\(n(n+1) / n^{\wedge} 2=37 / 36\)
\(n+1 / n=37 / 36\)
cross multiplication
\(36(n+1)=37 n\)
\(36 n+36=37 n\)
\(37 n-36 n=36\)
\(n=36\)
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