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The number of terms in the series $101+99+97+\ldots . .+47$ is
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The correct answer is:
$28$
Given series $101+99+97+\ldots \ldots \ldots .+47$
So, first term $a=101$, common difference $d=-2$ and last term $I=47$
We know that last term of a series
$T_1=a+(n-1) d \Rightarrow 47=101+(n-1)(-2)$
$\Rightarrow-54=(n-1)(-2) \Rightarrow n=28$
So, first term $a=101$, common difference $d=-2$ and last term $I=47$
We know that last term of a series
$T_1=a+(n-1) d \Rightarrow 47=101+(n-1)(-2)$
$\Rightarrow-54=(n-1)(-2) \Rightarrow n=28$
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