Question

Find the sum of all two digit natural number which are multiple of 4.

Answer 1

Answer:

1188

Step-by-step explanation:

first = 12

a=12

d=4

last digit=96

a. =a+(n-1)d

a. =12+(n-1)4

96=12+4n-4

96=8+4n

4n=96-8

4n=88

n=22

s. =n/2(2a+(n-1)d

s. =22/2(12*2+(22-1)4

s. =11(24+21*4)

s. =11(24+84)

s. =11(108)

s. =1188