Answer:
Step-by-step explanation:
What this question is asking of you is what is the greatest common divisor of 12 and 15. Or, what is the biggest number that divides both 12 and 15.
in order to find this we have to split each number into it's prime components.
for 12 they are 2,2 and 3 (
2
⋅
2
⋅
3
=
12
)
and for 15 they are 3 and 5 (
3
⋅
5
=
15
)
Out of those two groups (2,2,3) and (3,5) the only thing in common is 3, so 3 is the greatest common divisor. That tells us that the greatest number of groups that can exist and have the same number of girls and the same number of boys for each group is 3.
Now to find out how many girls and boys there are going to be in each group we divide the totals by 3, so:
12
3
=
4
girls per group, and
15
3
=
5
boys per group.
(just as a thought exercise, if there were 16 boys, the divisors would have been (2,2,3) and (2,2,2,2), leaving us with 4 groups [
2
⋅
2
] of 3 girls [12/4] and 4 boys [16/4] )
Answer:
9
Step-by-step explanation:
Composite Numbers before 10: 4, 6, 8, and 9
The only one of those 4 that is NOT a multiple of 2: 9
Given that
and
, if
is an arithmetic sequence, then the common difference between successive terms is
.
You then have
So the explicit formula for the
th term is
Answer:
100
Step-by-step explanation:
We have the sum of first n terms of an AP,
Sn = n/2 [2a+(n−1)d]
Given,
36= 6/2 [2a+(6−1)d]
12=2a+5d ---------(1)
256= 16/2 [2a+(16−1)d]
32=2a+15d ---------(2)
Subtracting, (1) from (2)
32−12=2a+15d−(2a+5d)
20=10d ⟹d=2
Substituting for d in (1),
12=2a+5(2)=2(a+5)
6=a+5 ⟹a=1
∴ The sum of first 10 terms of an AP,
S10 = 10/2 [2(1)+(10−1)2]
S10 =5[2+18]
S10 =100
This is the sum of the first 10 terms.
Hope it will help.
