Question

find thwe sum of series ,4+12+20+28.....+100.

Answer 1

a ( first term ) :- 4

d ( common difference ) :- 8

an = 100

n = ?

Sn = ?

an = a + ( n - 1 )d

100 = 4 + ( n - 1 ) 8

96 = 8n - 8

104 = 8n

13 = n



$sn = \frac{n}{2} (a + an)$


$sn = \frac{13}{2} (4 + 100)$


$sn = \frac{13}{2} (104)$
Sn = 13 × 52

Sn = 676

Answer 2

4+12+20+28.....+100,=676