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8. Sequences and Series
medium
Maximum value of sum of arithmetic progression $50, 48, 46, 44 ........$ is :-
A
$325$
B
$648$
C
$652$
D
$650$
Solution
For maximum sum $\Rightarrow \mathrm{T}_{\mathrm{n}}=0$
$a+(n-1) d=0$
$\Rightarrow 50+(n-1)(-2)=0 \Rightarrow n=26$
So $S_{26}=\frac{26}{2}[2 \times 50+25 \times(-2)]=650$
Standard 11
Mathematics