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How many terms of the A.P. 27, 24, 21, .... should be taken so that their sum is zero?

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#### Solution

The given AP is 27, 24, 21, ..

First term of the AP = 27

Common difference = 24 − 27 = −3

Let the sum of the first *x* terms of the AP be 0.

Sum of first *x* terms = `x/2`[2×27+(x−1)(−3)]=0

⇒`x/2`[54+(−3x+3)]=0

⇒x(54−3x+3)=0

⇒x(57−3x)=0

Now, either *x* = 0 or 57 − 3*x* = 0.

Since the number of terms cannot be 0, x≠0.

∴ 57 − 3*x* = 0

⇒ 57 = 3*x*

⇒ *x* = 19

Thus, the sum of the first 19 terms of the AP is 0.

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