Maximum value of the sum of the arithmetic pr | vedclass

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(D) For the sum to be maximum,we consider terms until they become non-negative. Since the common difference $d = -2$ is negative,the sum increases as

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$650$

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(D) For the sum to be maximum,we consider terms until they become non-negative. Since the common difference $d = -2$ is negative,the sum increases as long as the terms are positive.Set $T_n \geq 0$:$a + (n - 1)d \geq 0$$50 + (n - 1)(-2) \geq 0$$50 - 2n + 2 \geq 0$$52 \geq 2n \Rightarrow n \leq 26$.Thus,the sum of the first $26$ terms is the maximum sum.$S_{26} = \frac{26}{2} [2(50) + (26 - 1)(-2)]$$S_{26} = 13 [100 - 50]$$S_{26} = 13 	imes 50 = 650$.

Maximum value of the sum of the arithmetic progression $50, 48, 46, 44, \dots$ is:

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