Class 11MathematicsSequences and Seriesnth term of special series, Sum to n terms and Infinite number of terms
MediumIf $\frac{2+4+6+8+\dots+\text{upto } n \text{ terms}}{1+3+5+7+\dots+\text{upto } n \text{ terms}} = \frac{37}{36}$,then $n = $
- A$36$
- B$29$
- C$23$
- D$37$
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Similar Questions
The odd numbers are divided as follows:
Row $1$: $1, 3$
Row $2$: $5, 7, 9, 11$
Row $3$: $13, 15, 17, 19, 21, 23$
Then the sum of the $n^{th}$ row is:
Row $1$: $1, 3$
Row $2$: $5, 7, 9, 11$
Row $3$: $13, 15, 17, 19, 21, 23$
Then the sum of the $n^{th}$ row is:
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View SolutionLet $A = \sum_{i=1}^{10} \sum_{j=1}^{10} \min \{i, j\}$ and $B = \sum_{i=1}^{10} \sum_{j=1}^{10} \max \{i, j\}$. Then $A + B$ is equal to
DifficultJEE MAIN 2022
View SolutionThe ${n^{th}}$ term of the series $\frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \dots$ is:
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View SolutionIf $S_n = 2 + 4 + 7 + 11 + \dots + n$ terms,then $t_n = \dots$
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