Maximum value of sum of arithmetic progressio | vedclass

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Question

For maximum sum $\Rightarrow \mathrm{T}_{\mathrm{n}}=0$

$a+(n-1) d=0$

$\Rightarrow 50+(n-1)(-2)=0 \Rightarrow n=26$

So $S_{26}=\frac{26}{2}[2

Vedclass · Accepted Answer

$650$

Vedclass · Answer

For maximum sum $\Rightarrow \mathrm{T}_{\mathrm{n}}=0$

$a+(n-1) d=0$

$\Rightarrow 50+(n-1)(-2)=0 \Rightarrow n=26$

So $S_{26}=\frac{26}{2}[2 	imes 50+25 	imes(-2)]=650$

Maximum value of sum of arithmetic progression $50, 48, 46, 44 ........$ is :-

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