Question

The sum of all numbers between 1 and 200 which are multiples of 4

Answer 1

a = 4

d = 4

l = $a_{n}$ = 196

$a_{n}$ = a + (n-1)d

196 = 4 + (n - 1)4

n - 1 = 192/4

n - 1 = 48

n = 49

$S_{n}$ = n/2(a + $a_{n}$)

=49/2 ( 4 + 196)

= 49/2 (200)

= 49(100)

= 4900

Answer 2

a=4
d=4
tn= a+(n- 1)d
196=4+(n- 1)4
n- 1=192/4
n -1 =48
n=49
Sn=n/2(a+t n)
49/2(4+196)
49/2(200)
49 ,200/2= 100
49(100 )
= 49 multiply 100= 4900