CAT 2004 — QA Question 5

Let a and d be the first term and the common difference of the AP.

Sum of the first n terms of this A.P. = $\frac{n}{2}$[2a + (n - 1)d]

∴ The sum of the first 30 terms = 15 × (2a + 29d) …(i)

By conditions,

$\frac{11}{2}$ × (2a + 10d) = $\frac{19}{2}$ × (2a + 18d)

∴ 11a + 55d = 19a + 171d

∴ 8a = –116d

∴ a = - $\frac{29d}{2}$

From (i),

∴ The sum of the first 30 terms = 15 × (–29d + 29d)

∴ The sum of the first 30 terms = 0

Hence, option (a).

Alternatively,

If the sum of the first p terms of an A.P. is equal to the sum of the first q terms of the A.P. such that p and q are different, then the sum of the first (p + q) terms of the A.P. is zero.

∴ As the sum of the first 11 and the sum of the first 19 terms of the A.P. is equal, the sum of the first (11 + 19) = 30 terms of the A.P. is zero.

Hence, option (a).