Question

In an arithmetic progression, the 4th term equals three times the first term and the 7th term exceeds two times the third term by one. The sum of its first ten terms is:

• A. 100
• B. 108
• C. 120
• D. 124

Hint:

In problems like these, use the nth term formula and given conditions to set up equations. Then, solve for the first term and common difference. Finally, use the sum formula to find the desired value.

Answer 1

The Correct Option is C

In an arithmetic progression, the nth term is given by the formula: \[ T_n = a + (n-1) d \] where \(a\) is the first term and \(d\) is the common difference. Let's use the given conditions: 1. \(T_4 = 3 \cdot T_1 \implies a + 3d = 3a \implies 3d = 2a \implies d = \frac{2a}{3}\) 2. \(T_7 = 2 \cdot T_3 + 1 \implies a + 6d = 2(a + 2d) + 1 \implies a + 6d = 2a + 4d + 1 \implies 2d = a + 1 \implies d = \frac{a+1}{2}\) Now, solving for \(a\) and \(d\): From the equations \(d = \frac{2a}{3}\) and \(d = \frac{a+1}{2}\), we equate the two expressions: \[ \frac{2a}{3} = \frac{a+1}{2} \] Cross-multiply: \[ 4a = 3(a + 1) \implies 4a = 3a + 3 \implies a = 3 \] Substitute \(a = 3\) into \(d = \frac{2a}{3}\): \[ d = \frac{2 \times 3}{3} = 2 \] Now, the sum of the first 10 terms is: \[ S_{10} = \frac{10}{2} [2a + (10-1)d] = 5 [2 \times 3 + 9 \times 2] = 5 [6 + 18] = 5 \times 24 = 120 \] Thus, the sum of the first 10 terms is 120.