Discussion

Explanation:

Let there be n rows and a students in the first row.

∴ Number of students in the second row = a + 3

∴ Number of students in the third row = a + 6 and so on.

∴ The number of students in each row forms an arithmetic progression with common difference = 3

The total number of students = The sum of all terms in the arithmetic progression

n[2a+3(n-1)]2 = 30

Now consider options.

Option (a): n = 3

3[2a+3(3-1)]2 = 630

∴ a = 207

Option (b): n = 4

4[2a+3(4-1)]2 = 630

∴ a = 153

Option (c): n = 5

5[2a+3(5-1)]2 = 630

∴ a = 120

Option (d): n = 6

6[2a+3(6-1)]2 = 630

∴ a = 195/2 = 97.5

Option (e): n = 7

7[2a+3(7-1)]2 = 630

∴ a = 81

As a is an integer, only n = 6 is not possible.

Hence, option (d).

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