Question

How many terms of the A.P. 21, 18, 15, ... must be added to get the sum zero?

Hint:

For any AP with a positive first term \( a \) and a negative common difference \( d \), the sum becomes zero when the positive terms are perfectly balanced by the negative terms.
The middle term of such a symmetric sequence must be 0, which occurs here at the 8th term.
Thus, there are 7 positive terms, 1 zero term, and 7 negative terms, giving a total of \( 7 + 1 + 7 = 15 \) terms.

Answer 1

Step 1: Note the AP and what is being asked.
The AP is $21, 18, 15, \dots$, so first term $a = 21$ and common difference $d = 18 - 21 = -3$. We need the number of terms $n$ for which the sum of the first $n$ terms is zero.

Step 2: Write the nth term (the last term of this stretch) in terms of n.
\[ l = a_n = a + (n-1)d = 21 + (n-1)(-3) = 21 - 3n + 3 = 24 - 3n \]

Step 3: Use the sum formula built from the first and last term, not the 2a+(n-1)d form.
\[ S_n = \frac{n}{2}(a + l) \]
Substitute $a = 21$ and $l = 24 - 3n$:
\[ S_n = \frac{n}{2}(21 + 24 - 3n) = \frac{n}{2}(45 - 3n) \]

Step 4: Set this sum equal to zero and solve.
\[ \frac{n}{2}(45 - 3n) = 0 \]
Since $n$ is the number of terms, $n \neq 0$, so the factor $\frac{n}{2}$ cannot be zero. This forces the other factor to be zero:
\[ 45 - 3n = 0 \]
\[ 3n = 45 \]
\[ n = 15 \]

Final Answer:
\[ \boxed{n = 15} \]
15 terms of this AP must be added to get a sum of zero.