Your solution seems correct - but it seems u have a couple of syntax errors if you're using python You also have to re-initialize sum1 & sum2 between test cases. Right now you get the total sum of all arrs rather than arr[x]-hopefully that makes sense. You also have to remove duplicate numbers from the sum. For instance, if you're getting the sum of all numbers up to 20 - you'll end up adding 15 twice as it is divisible by 3 and 5. So the inner for loops will add it twice to the sum. So you'll need to remove 15 to get rid of duplicates.

for x in range(len(arr)): #by default range starts at o. therefore, range(len(arr)) = range(0, len(arr)) 
    sum1 = 0 # you forgot to initialize sum1
    sum2 = 0 # you forgot to initialize sum2
    duplicates = 0 #you have to remove duplicates from the answer
    for b5 in range (0, arr[x], 5): #you have to add the colons here
        sum1 = sum1 + b5
    for b3 in range (0, arr[x], 3): #you have to add the colons here
        sum2 = sum2 + b3
    for dup in range(0, arr[x], 3*5): # removes duplicates from the final sum
        duplicates = duplicates + dup
    sum = sum1 + sum2 - duplicates 
    print(sum)

This is an O(n^2) solution - you can drop it down to O(n) using a little bit of math.

You'll notice that the inner for loops can be represented using the formula sum(n)=Σd*i=d*Σi-where the summation starts at i = 0, end at ⌊(n-1)/d⌋ and d is the divisor (in the case of your question d=3 or 5).

for b5 in range (0, arr[x], 5):
    sum1 = sum1 + b5

(https://en.wikipedia.org/wiki/Summation)

There is a very common summation formula that is commonly used to convert summations into a closed-form expression (something with finite steps - which is O(1))

Σi=n*(n+1)/2

In the case of the inner loop - it would be sum(n) = d*(⌊(n-1)/d⌋)*(⌊(n-1)/d⌋+1)/2.

let,

f(n,d) = (⌊(n-1)/d⌋+1)/2

Therefore, the solution to your problem would be f(n,3)+f(n,5)-f(n,3*5)

Which would convert the inner for loops from O(n) to O(1). Which, means your entire solution would be O(n).

I'll let you figure out the code on your own. However, theoretically, there is a better solution; such that as arr grows indefinitely the work scales linearly rather than quadratically.

In Python you can iterate over the list itself:

for item in my_list:
   #do something with item

or to use indices you can use xrange():

for i in xrange(1,len(my_list)):    #as indexes start at zero so you 
                                    #may have to use xrange(len(my_list))
    #do something here my_list[i]

There's another built-in function called enumerate(), which returns both item and index:

for index,item in enumerate(my_list):
    # do something here

examples:

In [117]: my_lis=list('foobar')

In [118]: my_lis
Out[118]: ['f', 'o', 'o', 'b', 'a', 'r']

In [119]: for item in my_lis:
    print item
   .....:     
f
o
o
b
a
r

In [120]: for i in xrange(len(my_lis)):
    print my_lis[i]
   .....:     
f
o
o
b
a
r

In [122]: for index,item in enumerate(my_lis):
    print index,'-->',item
   .....:     
0 --> f
1 --> o
2 --> o
3 --> b
4 --> a
5 --> r

The length of an list can be get by len. Lists in a list can be traversed by:

for i in range(len(keys)):
    for j in range(len(keys[i])):

Another option is to use enumerate:

for i, keylist in enumerate(keys):
    for j, letter in enumerate(keylist):