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8. Sequences and Series
easy
The number of terms in the series $101 + 99 + 97 + ..... + 47$ is
A
$25$
B
$28$
C
$30$
D
$20$
Solution
(b) Given series $101 + 99 + 97 + ……… + 47$
So, first term $a = 101$, common difference $d = – 2$ and last term $l = 47$
We know that last term of a series
${T_l} = a + (n – 1)d$
$\Rightarrow 47 = 101 + (n – 1)( – 2)$
$ \Rightarrow $$ – 54 = (n – 1)( – 2) $
$\Rightarrow n = 28$.
Standard 11
Mathematics