Number of terms in series 101 + 99 + 97 + ..... + 47 is

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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

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