Given:
First term of an arithmetic sequence is 2.
Sum of first 15 terms = 292.5
To find:
The common difference.
Solution:
We have,
First term: 
Sum of first 15 terms: 
The formula of sum of first n terms of an AP 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)
Where, a is first term and d is common difference.
Putting , n=15 and a=2 in the above formula, we get
![292.5=\dfrac{15}{2}[2(2)+(15-1)d]](https://tex.z-dn.net/?f=292.5%3D%5Cdfrac%7B15%7D%7B2%7D%5B2%282%29%2B%2815-1%29d%5D)
![292.5=\dfrac{15}{2}[4+14d]](https://tex.z-dn.net/?f=292.5%3D%5Cdfrac%7B15%7D%7B2%7D%5B4%2B14d%5D)
![292.5=15[2+7d]](https://tex.z-dn.net/?f=292.5%3D15%5B2%2B7d%5D)
Divide both sides by 15.
Dividing both sides by 7, we get
Therefore, the common difference is 2.5.
Answer: D) cube root of 16
================================================
Explanation:
The rule we use is
![x^{m/n} = \sqrt[n]{x^m}](https://tex.z-dn.net/?f=x%5E%7Bm%2Fn%7D%20%3D%20%5Csqrt%5Bn%5D%7Bx%5Em%7D)
In this case, x = 4, m = 2 and n = 3.
So,
![x^{m/n} = \sqrt[n]{x^m}\\\\\\4^{2/3} = \sqrt[3]{4^2}\\\\\\4^{2/3} = \sqrt[3]{16}\\\\\\](https://tex.z-dn.net/?f=x%5E%7Bm%2Fn%7D%20%3D%20%5Csqrt%5Bn%5D%7Bx%5Em%7D%5C%5C%5C%5C%5C%5C4%5E%7B2%2F3%7D%20%3D%20%5Csqrt%5B3%5D%7B4%5E2%7D%5C%5C%5C%5C%5C%5C4%5E%7B2%2F3%7D%20%3D%20%5Csqrt%5B3%5D%7B16%7D%5C%5C%5C%5C%5C%5C)
Showing that the original expression turns into the cube root of 16.
For a translation 4 units down you would subtract 4 from each y value and 5 units right you would add 5 to each x value.
Your resulting points would be.
A'(-1,1), B'(-1,-2), C'(3,-2), D'(3,2)
