Substitute a = 18, d = -2, and S_n = 0 into the formula. This gives 0 = (n)/(2) [36 - 2n + 2], which simplifies inside the brackets to 38 - 2n.

how many times of an ap 181640 must be taken so that there sum is zero

Answer: $19$

Step-by-step solution

Step 1: Identify the first term and common difference

From the given arithmetic progression $18, 16, 14, \dots$ , the first term is $a = 18$ . The common difference is found by subtracting the first term from the second term, giving $d = 16 - 18 = -2$ .

Step 2: Apply the sum formula

We need the sum of the first $n$ terms to be $0$ , so we set $S_n = 0$ . We use the standard NCERT formula for the sum of the first $n$ terms of an AP when the last term is unknown.

Step 3: Substitute the values into the sum formula

Substitute $a = 18$ , $d = -2$ , and $S_n = 0$ into the formula. This gives $0 = \frac{n}{2} [36 - 2n + 2]$ , which simplifies inside the brackets to $38 - 2n$ .

Step 4: Simplify and solve for n

Dividing the terms inside the brackets by $2$ gives $n(19 - n) = 0$ . This yields two solutions: $n = 0$ or $19 - n = 0$ . Since the number of terms $n$ must be a positive integer, $n \neq 0$ .

Step 5: State the final value of n

Solving $19 - n = 0$ gives $n = 19$ . Therefore, taking $19$ terms of the given AP results in a sum of $0$ .

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