It is simple A.P problem. By mere inspection we get First term (a) of the series is 105 and last term (l) is 994 . By using the formula t th term = a + (n-1).d where n = no. of terms in series and d= difference which in this case is 7 ( since multiples of 7 is given ) . From this we get n = 128. Then using the sum formula of an A.P. series S(n) = (n/2 )*(a+l) we get the sum to be 70336.

1st term $a =105 = 7\times 15$

Last term $= 994 = 7\times 142$

Number of terms $n = 142-15 +1= 128$

Difference $d = 7$

$S_n= n/2 (( 2a+ (n-1) d)$

$S_n = 70336$

I have made it with c programming

include<stdio.h>

int a[10000]; int main() { int sum=0,i; for(i=100;i<1000;i++) { if((i%7)==0) { a[i]=i; } } for(i=100;i<1000;i++) { printf("%d\t",a[i]); sum=sum+a[i]; } printf("%d",sum); }

First term of multiple of 7 between 100 and 1000 = a = 105

Last term = L = 994

Total no. of terms = n = (994 - 105)/7 + 1 = 128

Difference between terms = d = 7

Sum of terms = S = n/2(2 a + (n-1) d) = 128/2(2 105 + (127) 7) = 64(210 + (889) = 70336

Therefore, sum of all terms of multiple of 7 between 100 and 1000 is 70,336.

it is an AR FIRST OF ALL. 142x7=994, &15x7=105, so use the formula of n(n+1)/2 from 1 to 142 then subtract the summation of first 14 natural no. and the multiply this with 7 as it is a multiple of 7 . and the answer is 70336,

c = 0
for x in range(100, 1000) :
if (x%7) == 0 :
c = c + x
print c

I personally was feeling incredibly time wasteful today, so I wrote a reusable version in my childhood programming language, Microsoft Small Basic. Took a hundred lines, but hey, it can write the whole list of multiples of any number over any range to a CSV file if so desired, as well as calculating the product and sum of the multiples. Twenty minutes of my life down the drain...

Subroutines for the sum, where LowerLimit = 100, UpperLimit = 1000, and Factor = 7:

Sub Create_Array
 Count = 1
  For i = LowerLimit To UpperLimit
    If Math.Remainder(i, Factor) = 0 Then
      ListOfMultiples[Count] = i
      Count = Count + 1
    EndIf
  EndFor
EndSub

Sub Sum
  Total = 0
  For i = 1 To Array.GetItemCount(ListOfMultiples)
    Total = Total + ListOfMultiples[i]
  EndFor
  TextWindow.WriteLine("The sum is " + Total + ".")
EndSub

Ugly code, unnecessary code, but by gar, it did the job, it's reusable, and it was fun.