Consecutive prime sum :Problem 50

Consecutive prime sum :Problem 50

The prime 41, can be written as the sum of six consecutive primes:

41 = 2 + 3 + 5 + 7 + 11 + 13

This is the longest sum of consecutive primes that adds to a prime below one-hundred.
The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.
Which prime, below one-million, can be written as the sum of the most consecutive primes?

Problem Source : Euler Project

Python Code 

import math
def largestSumofConsecutivePrime():
    def checkPrime(numb):
        prime = True
        for i in range(int(math.sqrt(numb))):
            i = i+2
            if numb%i == 0:
                prime = False
                break
        if numb == 2:
            return True
        else:           
            return prime
    primeList = ()   
    cumiSumList = ()
    sumd = 0
    count = 0
    cumiSumList =cumiSumList+(0,)
    maxPrime = 0
    maxLength = 0
    for prime in range(1000000):
        prime = prime + 2
        if checkPrime(prime) == True:
            primeList = primeList+(prime,)
            sumd = sumd + prime
            cumiSumList =cumiSumList+(sumd,)
            count = count + 1
    #print primeList
    #print cumiSumList
    for i in range(count):
        for j in range(i+1):
            if cumiSumList[i]-cumiSumList[j] > 1000000:   
                break
            if (cumiSumList[i]-cumiSumList[j]) in primeList:
                if i-j > maxLength:
                    maxLength = i-j
                    maxPrime = cumiSumList[i] - cumiSumList[j]
    print maxLength,maxPrime                   
largestSumofConsecutivePrime()