Find the sum of first 14 term of the sequence: -3, -1,1,3,...

SOLUTION:Step 1: List the given values.[tex]\begin{aligned} & n = 14 \\ & a_1 = -3 \\ & a_2 = -1 \\ & a_3 = 1 \\ & a_4 = 3 \end{aligned}[/tex]Step 2: Find the common difference.[tex]\begin{aligned} d & = a_2 - a_1 = a_3 - a_2 = a_4 - a_3 \\ d & = -1 - (-3) = 1 - (-1) = 3 - 1 \\ & = -1 + 3 = 1 + 1 = 3 - 1 \\ d & = 2 = 2 = 2 \end{aligned}[/tex]Step 3: Solve for the sum of the terms of an arithmetic sequence.[tex]\begin{aligned} S & = \frac{n}{2}[2a_1 + (n - 1)d] \\ & = \frac{14}{2}[2(-3) + (14 - 1)(2)] \\ & = 7[(-6 + (13)(2)] \\ & = 7(-6 + 26) \\ & = 7(20) \\ & = \boxed{140} \end{aligned}[/tex]Hence, the sum of the first 14 terms of the sequence is 140.

Answered by GreatRedSpot | 2024-09-01