Respuesta :
he will have 5 sets because 2 plus 5 plus 8 plus 11 plus 14 is 40
The number of marbles in each set is an arithmetic sequence. Â The sum of any arithmetic sequence is that average of the first and last terms times the number of terms...mathematically this is:
s(n)=(2an+dn^2-dn)/2, a=initial term, n=term number, d=common difference
In this case the first term is 2 and since each term is 3 more than the previous term, the common difference is 3, so now we can say:
s(n)=(2*2n+3n^2-3n)/2
s(n)=(3n^2+n)/2 Â and we are told that there is a total of 40 so
(3n^2+n)/2=40
3n^2+n=80
3n^2+n-80=0
3n^2-15n+16n-80=0
3n(n-5)+16(n-5)=0
(3n+16)(n-5)=0, since n>0
n=5
So Abe will make five sets of marbles...
(The sets contain 2,5,8,11, and 14 marbles, which sum to 40 marbles total)
s(n)=(2an+dn^2-dn)/2, a=initial term, n=term number, d=common difference
In this case the first term is 2 and since each term is 3 more than the previous term, the common difference is 3, so now we can say:
s(n)=(2*2n+3n^2-3n)/2
s(n)=(3n^2+n)/2 Â and we are told that there is a total of 40 so
(3n^2+n)/2=40
3n^2+n=80
3n^2+n-80=0
3n^2-15n+16n-80=0
3n(n-5)+16(n-5)=0
(3n+16)(n-5)=0, since n>0
n=5
So Abe will make five sets of marbles...
(The sets contain 2,5,8,11, and 14 marbles, which sum to 40 marbles total)