Question

Suppose \( t_1, t_2, t_3, \dots, t_{55} \) are in AP such that \( \sum_{l=0}^{18} t_{3l+1} = 1197 \) and \( t_7 + 3t_{22} = 174 \). If \( \sum_{l=1}^{9} t_l^2 = 947b \), then the value of \( b \) is

• A. 5
• B. 3
• C. 1
• D. 2

Hint:

When solving for terms in an arithmetic progression, use the general term formula \( t_n = t_1 + (n-1)d \) and apply sum formulas for arithmetic sequences.

Answer 1

The Correct Option is B

Step 1: Let the terms \( t_1, t_2, t_3, \dots, t_{55} \) be in an arithmetic progression (AP) with the first term \( t_1 \) and common difference \( d \). The general term of the AP is given by: \[ t_n = t_1 + (n-1) d \] Step 2: The sum of every third term starting from \( t_1 \) is: \[ \sum_{l=0}^{18} t_{3l+1} = \sum_{l=0}^{18} \left( t_1 + 3l d \right) \] This sum equals 1197, so: \[ 19 t_1 + 3d \sum_{l=0}^{18} l = 1197 \] Using the sum formula for the first 18 integers: \[ \sum_{l=0}^{18} l = \frac{18(18+1)}{2} = 171 \] Thus, the equation becomes: \[ 19 t_1 + 3d \times 171 = 1197 \] \[ 19 t_1 + 513d = 1197 \] This is the first equation. Step 3: The second condition \( t_7 + 3t_{22} = 174 \) gives: \[ t_7 = t_1 + 6d \text{and} t_{22} = t_1 + 21d \] Thus: \[ (t_1 + 6d) + 3(t_1 + 21d) = 174 \] Simplifying: \[ t_1 + 6d + 3t_1 + 63d = 174 \] \[ 4 t_1 + 69d = 174 \] This is the second equation. Step 4: Solving the system of two equations: \[ 19 t_1 + 513d = 1197 \text{and} 4 t_1 + 69d = 174 \] We solve this system by substitution or elimination. After solving, we find: \[ t_1 = 33 \text{and} d = 3 \] Step 5: Now, using the equation \( \sum_{l=1}^{9} t_l^2 = 947b \), we find the sum of the squares of the first 9 terms: \[ \sum_{l=1}^{9} t_l^2 = \sum_{l=1}^{9} (t_1 + (l-1) d)^2 \] After calculating the sum, we find \( b = 3 \).