CAT 2006 — QA Question 6

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

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

Now consider options.

Option (a): n = 3

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

∴ a = 207

Option (b): n = 4

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

∴ a = 153

Option (c): n = 5

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

∴ a = 120

Option (d): n = 6

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

∴ a = 195/2 = 97.5

Option (e): n = 7

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

∴ a = 81

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

Hence, option (d).