The maximum value of the sum of the Arithmeti | vedclass

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(D) For the sum to be maximum,we consider terms until they become non-negative. Let the $n^{th}$ term be $T_n = a + (n-1)d$. Here,$a = 50$ and $d = -2

Vedclass · Accepted Answer

$650$

Vedclass · Answer

(D) For the sum to be maximum,we consider terms until they become non-negative. Let the $n^{th}$ term be $T_n = a + (n-1)d$. Here,$a = 50$ and $d = -2$. $T_n = 50 + (n-1)(-2) = 50 - 2n + 2 = 52 - 2n$. Setting $T_n \geq 0$,we get $52 - 2n \geq 0$ $\Rightarrow 2n \leq 52$ $\Rightarrow n \leq 26$. Thus,the sum is maximum for $n = 26$. $S_{26} = \frac{26}{2} [2(50) + (26-1)(-2)] = 13 [100 - 50] = 13 	imes 50 = 650$.

The maximum value of the sum of the Arithmetic Progression $50, 48, 46, 44, \dots$ is:

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