Step-by-step Solution:

Correct Answer: C $(3)$ To find the value of \(b\), we first determine the first term \(t_1\) and the common difference \(d\) of the Arithmetic Progression (AP) using the first two conditions. Step 1: Find the First Term \(t_1\) and Common Difference \(d\) Using the first condition: \(\sum_{i=0}^{18} t_{3i+1} = 1197\) The terms in this summation \(t_1, t_4, t_7, \dots, t_{55}\) form an AP with: \[ A=t_1,\qquad D=t_4-t_1=(t_1+3d)-t_1=3d,\qquad N=18-0+1=19. \] Using the sum formula \(S_N=\frac{N}{2}\left(2A+(N-1)D\right)\), \[ 1197=\frac{19}{2}\left(2t_1+(19-1)(3d)\right) \] \[ 1197=\frac{19}{2}\left(2t_1+18\cdot 3d\right)=19(t_1+27d). \] \[ t_1+27d=\frac{1197}{19}=63 \qquad \text{(Equation 1)} \] Using the second condition: \(t_7+3t_{22}=174\) \[ t_7=t_1+6d,\qquad t_{22}=t_1+21d \] \[ (t_1+6d)+3(t_1+21d)=174 \] \[ t_1+6d+3t_1+63d=174 \] \[ 4t_1+69d=174 \qquad \text{(Equation 2)} \] Solving the system: From Equation 1, \[ t_1=63-27d. \] Substitute into Equation 2: \[ 4(63-27d)+69d=174 \] \[ 252-108d+69d=174 \] \[ 252-39d=174 \] \[ 39d=78 \] \[ d=2. \] Then from Equation 1, \[ t_1=63-27(2)=63-54=9. \] So, \(t_1=9\) and \(d=2\). Step 2: Calculate the Sum of Squares Given: \(\sum_{i=1}^{9} t_i^2 = 947b\) General term: \[ t_i=t_1+(i-1)d=9+(i-1)\cdot 2=2i+7. \] So, \[ \sum_{i=1}^{9} t_i^2=\sum_{i=1}^{9}(2i+7)^2 =\sum_{i=1}^{9}(4i^2+28i+49). \] Using \(\sum_{i=1}^{9} i = 45\) and \(\sum_{i=1}^{9} i^2 = 285\), \[ \sum_{i=1}^{9}(4i^2+28i+49)=4\sum_{i=1}^{9}i^2+28\sum_{i=1}^{9}i+49\cdot 9 \] \[ =4(285)+28(45)+441 \] \[ =1140+1260+441=2841. \] Step 3: Find the Value of \(b\) \[ 2841=947b \] \[ b=\frac{2841}{947}=3. \]