Maximum value of the sum of the Arithmetic Pr | vedclass

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(D) For the sum of an Arithmetic Progression to be maximum, we consider terms until they become non-negative.Given sequence: $50, 48, 46, 44, \dots$He

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$650$

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(D) For the sum of an Arithmetic Progression to be maximum, we consider terms until they become non-negative.Given sequence: $50, 48, 46, 44, \dots$Here, first term $a = 50$ and common difference $d = 48 - 50 = -2$.The $n$-th term is given by $T_n = a + (n - 1)d$.To find the number of positive terms, we set $T_n > 0$:$50 + (n - 1)(-2) > 0$$50 - 2n + 2 > 0$$52 > 2n \Rightarrow n Thus, there are $25$ positive terms. The $26$-th term is $50 + (26 - 1)(-2) = 50 - 50 = 0$.Adding $0$ does not change the sum, so the sum of the first $25$ terms is equal to the sum of the first $26$ terms.Using the sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$:$S_{26} = \frac{26}{2}[2(50) + (26 - 1)(-2)]$$S_{26} = 13[100 - 50] = 13 	imes 50 = 650$.

Maximum value of the sum of the Arithmetic Progression $50, 48, 46, 44, \dots$ is

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