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What is the sum of the following series:

$$2 + 4+ 6+ 8+ \dots + 94 + 96 + 98 + 100$$

  1. $4300$
  2. $2000$
  3. $3240$
  4. $2550$
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Given  that $S=2+4+6+8+10+.....+94+96+98+100$

This is the arithmetic progression(series)

first term $a=2,$Common difference $d=6-4=4-2=2$

Last term $l=100$

Now,we can find the total number of terms $n=?$

        $l=a+(n-1)d$                       $n^{th}$  $term$ $T_{n}=a+(n-1)d$

        $100=2+(n-1)2$

         $n=50$

Now,$S=\frac{n}{2}\left[a+l\right]$

         $S=\frac{50}{2}\left[2+100\right]$

          $S=\frac{50}{2}\left[102\right]$

          $S=2550$

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take common 2

2(1+2+3+4+..............+50)

2*(50*51/2)

50*51

2550
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