Hey,I don’t think this is the whole question
The question is asking you to solve for A
to determine the first, fourth and tenth term you simply put those values in for (n)
the first term you then be given by the equation A(1) = 12 + ((1)-1)(3)
A= 12
Same thing for 4th term A(4) = 12 + ((4)-1)(3)
A= 21
10th term: A(10) = 12 + ((10)-1)(3)
A= 39
∴ your answer is C
Ok so
x - the smallest
The other 4 are : x+1, x+2, x+3, x+4
We know that sum is 120
So
x+x+1+x+2+x+3+x+4=120
Combine like terms
5x+10=120
Subtract 10 on both sides
5x=120-10
5x=110
Divide by 5 on both sides
X=22
The third number is x+2 = 22+2=24
So in distributive property you multiply the outside number by the two inside so in this case 2(a) + 2(-2) = 3(a) + 3(-5)
Which will give you
2a+ - 4 = 3a + - 15
Or
2a-4=3a-15
Then if your solving for a your can subtract 3a from 2a giving you
- a-4=-15
Then add 4 to - 15
-a =-11
Then divide by negative 1 (this is so a isn't negative)
A=11
Hope this helps!