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Question:

A group of 630 children is arranged in rows for a group photograph session. Each row contains three fewer children than the row in front of it. What number of rows is not possible?

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

= $\frac{n [2 a + 3 (n - 1)]}{2}$ = 30

Now consider options.

Option (a): n = 3

$\frac{3 [2 a + 3 (3 - 1)]}{2}$ = 630

∴ a = 207

Option (b): n = 4

$\frac{4 [2 a + 3 (4 - 1)]}{2}$ = 630

∴ a = 153

Option (c): n = 5

$\frac{5 [2 a + 3 (5 - 1)]}{2}$ = 630

∴ a = 120

Option (d): n = 6

$\frac{6 [2 a + 3 (6 - 1)]}{2}$ = 630

∴ a = 195/2 = 97.5

Option (e): n = 7

$\frac{7 [2 a + 3 (7 - 1)]}{2}$ = 630

∴ a = 81

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

Hence, option (d).

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