Assertion (A): The sum of the first fifteen terms of the AP $21, 18, 15, 12, \dots$ is zero.
Reason (R): The sum of the first $n$ terms of an AP with first term $a$ and common difference $d$ is given by: $S_n = \frac{n}{2} \left[ a + (n - 1) d \right].$
For the AP: $a = 21$, $d = 18 - 21 = -3$. Sum of 15 terms:
\[S_n = \frac{n}{2} \left[ 2a + (n - 1)d \right]\]
Substitute $n = 15$, $a = 21$, $d = -3$:
\[S_{15} = \frac{15}{2} \left[ 2(21) + (15 - 1)(-3) \right] = \frac{15}{2} \left[ 42 - 42 \right] = 0\]
Both (A) and (R) are true, and (R) explains (A).
The remainder when \( 64^{64} \) is divided by 7 is equal to: