(A) The sum of the first $n$ even natural numbers is given by $S_e = n(n+1)$.The sum of the first $n$ odd natural numbers is given by $S_o = n^2$.Give

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(A) The sum of the first $n$ even natural numbers is given by $S_e = n(n+1)$.The sum of the first $n$ odd natural numbers is given by $S_o = n^2$.Given the ratio: $\frac{n(n+1)}{n^2} = \frac{37}{36}$.Simplifying the expression: $\frac{n+1}{n} = \frac{37}{36}$.By cross-multiplication: $36(n+1) = 37n$.$36n + 36 = 37n$.$37n - 36n = 36$.Therefore,$n = 36$.

Class 11 Mathematics Sequences and Series nth term of special series, Sum to n terms and Infinite number of terms

Medium

If $\frac{2+4+6+8+\dots+\text{upto } n \text{ terms}}{1+3+5+7+\dots+\text{upto } n \text{ terms}} = \frac{37}{36}$,then $n = $

MHT CET 2020 All MHT CET

A
$36$
B
$29$
C
$23$
D
$37$

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nth term of special series, Sum to n terms and Infinite number of terms

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If $\frac{2+4+6+8+\dots+\text{upto } n \text{ terms}}{1+3+5+7+\dots+\text{upto } n \text{ terms}} = \frac{37}{36}$,then $n = $

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