Find the sum of all two digit natural number which are multiple of 4.
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10,11,12,.....,99 are two digit natural numbers
12, 16,20 ,.....96 are two digit numbers which are divisible by 4
12 ,16, 20 ,....,96 are in AP
first term = a=12
common difference =d = a2-a1 = 16-12 =4
last term = a+(n-1)d= 96
12+(n-1)4= 96
(n-1)4= 96-12
(n-1)4 =84
(n-1)=84/4
n-1=21
n=21+1
n=22
a=4 , d=4 , n= 22
sn =n/2[2a+(n-1)d]
= 22/2[2*4+(22-1)4]
= 11[8+21*4]
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