help in maths_series and sequences?

Question

find the sum of all natural nums between 50 and 500 divisible by 7

Answer 1

17696

a(1) = 56
a(n) = 497
a(n) = a(1) + (n-1)d, so (n-1)d=441 and d=7 so n=64

S(n) = n(a(1)+a(n))/2 = 64(56+497)/2 = 32(553) = 17696

(Gee JinX, that wasnt your first answer....and there are 64 members in the series)

Answer 2

The natural numbers between 50 and 500 that are divisible by 7 start with 56 aand extend to 497.
Their sum is (56 + 63 + .... + 490 + 497)
which is 7(8 + 9 + .....+ 70 + 71)

This is 7(1+ 2.+ .. + 70+ 71) - 7(1+ 2.+ .. + 6+ 7)
which is 7 (71)(72)/2 - 7(7)(8)/2 = 17696

The desired sum is 17,696

Answer 3

first think about what this is asking
what is the first number greater than 50 divisible by 7?

56, right

what is the next one, 63, then 70, each time adding 7

now, you could go through and write them all down and add them up yourself or use a computer with a spreadsheet or something and just power it out, but that would defeat the point of this study of sequences and series

now consider the summation of 56+7n with n starting a 0, and going high enough to cover the range 500/7=71.4, and 71*7= 497 which you can see then is the last number in that range that is divisable by 7

497-56=441 441/7=63

so, Sum (56+7n) n from 0 to 63 will get all the terms

that is the set up, now you just have to look up the formula and crank it out

enjoy

Answer 4

17696

50 < 7 * 8 < 7 * 71 < 500
s = 7 * (8 + 9 + ... + 70 + 71)
s = 7 * ((1 + 2 + .. +71) - (1 + 2 + ... +7))
s = (7 * (72 * 71 - 7 * 8)) / 2
s = 17696

Alternative answer:
consider a(n) = a(n - 1) + 7; a(1) = 56 and calculate a sum of 63 members of this arithmetical progression with the formula

Answer 5

Gotta add a little Java in this:

class answer {
public static void main(String[] args){
int sum = 0;
int i;
for (i = 50; i < 501; i++){
if (i % 7 == 0){
sum += i;
}
}
System.out.println("Sum: " + sum);
}

}


Sum: 17696.