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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 − 3x = 0.
Since the number of terms cannot be 0, x≠0.
∴ 57 − 3x = 0
⇒ 57 = 3x
⇒ x = 19
Thus, the sum of the first 19 terms of the AP is 0.
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