Find the sum of all three digits multiples of 4.
class 10 cbse arthmetic progressions.

First 3 digit multiple = 100

Last 3 digit multiple = 996

AP: 100, 104, 108...996

an = 996

a + (n-1)d = 996

100 + (n-1)4 = 996

(n-1)4 = 896

(n-1) = 224

Therefore n = 225

Sn = n(a+l)/2

= 225(100+996)/2

= 225 x 548

= 123300

Hope this helps!

im not quite sure if this is right but i got 5.6 oisutuve

step-by-step explanation:

answer: (1,3) and (-2,0) i think.