The correct choice is D.) No solution
Answer:
![\sum_{n=1}^{25}(3n-2)=925]](https://tex.z-dn.net/?f=%5Csum_%7Bn%3D1%7D%5E%7B25%7D%283n-2%29%3D925%5D)
Step-by-step explanation:
The given series is 
The first term of this series is



The second term is



The common difference is

The sum of the first n-terms is given by;
![S_n=\frac{n}{2}[2a_1+d(n-1)]](https://tex.z-dn.net/?f=S_n%3D%5Cfrac%7Bn%7D%7B2%7D%5B2a_1%2Bd%28n-1%29%5D)
The sum of the first 25 terms of the series is
![S_{25}=\frac{25}{2}[2(1)+3(25-1)]](https://tex.z-dn.net/?f=S_%7B25%7D%3D%5Cfrac%7B25%7D%7B2%7D%5B2%281%29%2B3%2825-1%29%5D)
![S_{25}=\frac{25}{2}[2+3(24)]](https://tex.z-dn.net/?f=S_%7B25%7D%3D%5Cfrac%7B25%7D%7B2%7D%5B2%2B3%2824%29%5D)
![S_{25}=\frac{25}{2}(74)]](https://tex.z-dn.net/?f=S_%7B25%7D%3D%5Cfrac%7B25%7D%7B2%7D%2874%29%5D)
![S_{25}=925]](https://tex.z-dn.net/?f=S_%7B25%7D%3D925%5D)
<span>z2-z(z+3)+3z
=> z2 - z2 - 3z + 3z
=> 0
we can cancel the like terms if they have different signs ..
Hope it helps !!</span>