Find the sam of 1st 24term of the arithmetic sequence 14,32,50,68,86
Answer (1)
SOLUTION:Step 1: List the given values.[tex]\begin{aligned} & n = 24 \\ & a_1 = 14 \\ & a_2 = 32 \\ & a_3 = 50 \\ & a_4 = 68 \\ & a_5 = 86 \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 = a_5 - a_4 \\ d & = 32 - 14 = 50 - 32 = 68 - 50 = 86 - 68 \\ d & = 18 = 18 = 18 = 18 \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{24}{2}[2(14) + (24 - 1)(18)] \\ & = 12[28 + (23)(18)] \\ & = 12(28 + 414) \\ & = 12(442) \\ & = \boxed{5,304} \end{aligned}[/tex]Hence, the sum of the first 24 terms of the sequence is 5,304.
