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 must include all positive terms. The sum starts decreasing as soon as we add negative t

Vedclass · Accepted Answer

$650$

Vedclass · Answer

(D) For the sum of an arithmetic progression to be maximum, we must include all positive terms. The sum starts decreasing as soon as we add negative terms.Set the $n^{th}$ term $T_n = 0$ or find the last positive term.Given $a = 50$ and $d = -2$.$T_n = a + (n - 1)d = 50 + (n - 1)(-2) = 50 - 2n + 2 = 52 - 2n$.Setting $T_n > 0$, we get $52 - 2n > 0$, which implies $n Thus, the sum is maximum for $n = 26$ (where $T_{26} = 0$).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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