Question

The 6th term of an arithmetic sequence is 24 and the 8th term in 34. What is the sum of the first 13 terms of the arithmetic sequence?

• A. 277
• B. 297
• C. 377
• D. 397

Answer 1

Answer:

C. 377

Arithmetic Sequences and Series

This problem involves an arithmetic sequence, which is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference, denoted by d. The general form of an arithmetic sequence is a, a+d, a+2d, a+3d, ..., where a is the first term.

Key Formulas for Arithmetic Sequences:

• The n^th term of an arithmetic sequence is given by: a_n = a + (n-1)d

• The sum of the first n terms of an arithmetic sequence is given by: S_n = n/2 × [2a + (n-1)d] or S_n = n/2 × (a + a_n)

Problem Analysis:

• We are given that the 6th term (a₆) is 24.

• We are also given that the 8th term (a₈) is 34.

• We need to find the sum of the first 13 terms (S₁₃).

Step-by-Step Solution:

1. Finding the common difference (d):

   • Using the formula a_n = a + (n-1)d, we can write:

     • a₆ = a + (6-1)d = a + 5d = 24

     • a₈ = a + (8-1)d = a + 7d = 34

   • Subtracting the first equation from the second:

     (a + 7d) - (a + 5d) = 34 - 24

     2d = 10

     d = 5

2. Finding the first term (a):

   • Substitute the value of d (which is 5) into the equation for a₆:

     a + 5d = 24

     a + 5(5) = 24

     a + 25 = 24

     a = -1

3. Finding the sum of the first 13 terms (S₁₃):

   • Using the formula S_n = n/2 * [2a + (n-1)d]:

     S₁₃ = 13/2 × [2(-1) + (13-1)5]

     S₁₃ = 13/2 × [-2 + (12)5]

     S₁₃ = 13/2 × [-2 + 60]

     S₁₃ = 13/2 × [58]

     S₁₃ = 13 × 29 = 377