Given:
The first two terms in an arithmetic progression are -2 and 5.
The last term in the progression is the only number in the progression that is greater than 200.
To find:
The sum of all the terms in the progression.
Solution:
We have,
First term : 
Common difference : 


nth term of an A.P. is

where, a is first term and d is common difference.

According to the equation,
.



Divide both sides by 7.

Add 1 on both sides.

So, least possible integer value is 30. It means, A.P. has 30 term.
Sum of n terms of an A.P. is
![S_n=\dfrac{n}{2}[2a+(n-1)d]](https://tex.z-dn.net/?f=S_n%3D%5Cdfrac%7Bn%7D%7B2%7D%5B2a%2B%28n-1%29d%5D)
Substituting n=30, a=-2 and d=7, we get
![S_{30}=\dfrac{30}{2}[2(-2)+(30-1)7]](https://tex.z-dn.net/?f=S_%7B30%7D%3D%5Cdfrac%7B30%7D%7B2%7D%5B2%28-2%29%2B%2830-1%297%5D)
![S_{30}=15[-4+(29)7]](https://tex.z-dn.net/?f=S_%7B30%7D%3D15%5B-4%2B%2829%297%5D)
![S_{30}=15[-4+203]](https://tex.z-dn.net/?f=S_%7B30%7D%3D15%5B-4%2B203%5D)


Therefore, the sum of all the terms in the progression is 2985.
1/3 x - 1/2 = 18 1/2
1/3 x = 19 (add 1/2 on the right side to 18 1/2)
x = 57 (multiply the reciprocal of 1/3 and that will be 3/1 or 3 to 19 to get x by itself)
So, the answer is x = 57 (d. 57)
That would simplify to 32x^5+400x^4+200x^3+5000x^2+6250x+3125
You would do that because you have to write it repeatedly as (2x+5)(2x+5) and so on.
We want to know the time, <em>t</em>, it takes the ball to reach a height (<em>y</em>) of 0.

We can factor out the GCF first. The largest number that will divide evenly into 16 and 24 is 8. Also, both terms have a <em>t</em>, so we can factor that out as well:

(-16/8 = -2 and 24/8 = 3)
Using the zero product property, we know that either 8t=0 or -2t+3=0. Solving the first equation, we would divide both sides by 8:
8t/8=0/8
t=0
This is at 0 seconds, before the ball is in the air at all.
Solving the second equation, we start by subtracting 3 from both sides:
-2t+3-3=0-3
-2t=-3
Now we divide both sides by -2
-2t/-2=-3/-2
t=1.5
After 1.5 seconds, the ball will hit the ground again.