The '2 multiple' there are 50 terms, because $\frac{100}{2} = 50$ . Then $a$ is the first term and $b$ is the difference between each term.

$S_{50} = \frac{n}{2}(2a + (n-1)b)$

$S_{50} = \frac{50}{2}(2 \times 2 + (50-1)2)$

$S_{50} = 50(2+ 49$

$S_{50} = 2550$

Then the term that can divided by 5 but can't be divided by 2 because the terms have been included in the '2 multiple' terms.

$5+15+25+35+45+55+65+75+85+95= 500$

So, $2550+500=\boxed{3050}$

2(1+2+3+4+...+50) = 2550

5(1+2+3+...+20) = 1050

(multiples of 2 & 5) 10(1+2+...+10) = 550

Apply Inclusion-Exclusion Principle, i.e., 2550+1050-550=3050.

Thus, the answer is 3050.

include<stdio.h>

void main()  {
int i,j,k,sum;
int a[100];
for(i=2,k=0;i<=100;i++) {
    if(i % 2== 0 || i%5==0) {

        a[k]=i;
        k++;
    }
}
for(j=sum=0;j<k;j++)
    sum+=a[j];
    printf("\n\n%d,",sum);
}

S = 2 + 4 + 5 + 6 + 8 + 10 + 12 + 14 + 15 + 16 + .........+100 $$ = (2 + 4 + 6 + 8 + 10 + ..... + 100) + (5 + 15 + 25 + ..... + 95) $$ = 2x(1+2+3+4+.....+50) + 5(1+3+5+......+19) $$ = (2x½ x 50 x 51) + {5x½x(1+19)x10} $$ = 2550 + 500 $$ = 3050

sum of integers from 1 to 100 that are divisible by 2 =n(n+1) =50*51 =2550 sum of integers from 1 to 100 that are divisible by 5 but not divisible by 2. are 5,15,.25,----------95 = 10/2 (5+95) = 500 The sum of integers from 1 to 100 that are divisible by 2 or 5 is=2550+500=3050 Ans

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var sum=0;
for (var i=1;i<=100;i++) 
{
 if( (i%2==0) || (i%5==0) ) 
         sum+=i; 
 } 
console.log(sum); 

JavaScript code ( can run in browser console )

First sum digits divisible by 2 which is 2550. Then sum the digits divisible by 5 which is 1050. now subtract digits common to both which are digits divisible by 10 which is 550. therefore answer is 2550+1050-550=3050.

sum to 50 even numbers(since there are only 50 even numbers from 1 to 100) is 2550 [ using S_50 = 50/2 (2x2 + 49x2) ]
sum of numbers which are not divisible by 5 but divisible by 2 is
2550 - 10 (1+2+3+5+6+7+8+9+10) = 2000
the number of numbers divisible by 5 from 1 to 100 is 20. The sum of such numbers is (using the same formula above used) is 1050. Hence the sum of numbers that are divisible by 2 and 5 from 1 to 100 is 2000+1050 = 3050