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In which row does 2020 appear in the given figure?

 

\(\begin{array} {rrrrrrrrr} \text{Row 1:} & 1 \\ \text{Row 2:} & 2 & 3 \\ \text{Row 3:} & 4 & 5 & 6 \\ \text{Row 4:} & 7 & 8 & 9 & 10 \\ \text{Row 5:} & 11 & 12 & 13 & 14 & 15 \\ \cdot : & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot : & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot : & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \end{array}\)

 Dec 20, 2019

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 #1
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1/2 * n*(n + 1) =2016 [since 2016 is a perfect "triangular" number]

 

n = 63rd row. This means that the 63rd row will end in 2016.

 

Therefore, 2020 will appear in the next row, or the 64th row.

 Dec 20, 2019
 #1
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+1
Best Answer

1/2 * n*(n + 1) =2016 [since 2016 is a perfect "triangular" number]

 

n = 63rd row. This means that the 63rd row will end in 2016.

 

Therefore, 2020 will appear in the next row, or the 64th row.

Guest Dec 20, 2019

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