Insert five arithmetic means between -4 and 8.0​

SOLUTION:Step 1: List the given values.Since we must insert 5 arithmetic means between -4 and 8, the number of terms (5 arithmetic means + 2 given terms) is 7.[tex]\begin{aligned} & a_1 = -4 \\ & a_7 = 8 \\ & n = 7 \end{aligned}[/tex]Step 2: Determine the common difference.[tex]\begin{aligned} a_n & = a_1 + (n - 1)d \\ a_7 & = -4 + (7 - 1)d \\ 8 & = -4 + 6d \\ -4 + 6d & = 8 \\ 6d & = 8 + 4 \\ 6d & = 12 \\ \frac{6d}{6} & = \frac{12}{6} \\ d & = 2 \end{aligned}[/tex]Step 3: Solve for the 5 arithmetic means.• For a₂[tex]\begin{aligned} a_2 & = a_1 + (2 - 1)d \\ a_2 & = a_1 + d \\ a_2 & = -4 + 2 \\ & = \boxed{-2} \end{aligned}[/tex]• For a₃[tex]\begin{aligned} a_3 & = a_1 + (3 - 1)d \\ a_3 & = a_1 + 2d \\ a_3 & = -4 + 2(2) \\ a_3 & = -4 + 4 \\ & = \boxed{0} \end{aligned}[/tex]• For a₄[tex]\begin{aligned} a_4 & = a_1 + (4 - 1)d \\ a_4 & = a_1 + 3d \\ a_4 & = -4 + 3(2) \\ a_4 & = -4 + 6 \\ & = \boxed{2} \end{aligned}[/tex]• For a₅[tex]\begin{aligned} a_5 & = a_1 + (5 - 1)d \\ a_5 & = a_1 + 4d \\ a_5 & = -4 + 4(2) \\ a_5 & = -4 + 8 \\ & = \boxed{4} \end{aligned}[/tex]• For a₆[tex]\begin{aligned} a_6 & = a_1 + (6 - 1)d \\ a_6 & = a_1 + 5d \\ a_6 & = -4 + 5(2) \\ a_6 & = -4 + 10 \\ & = \boxed{6} \end{aligned}[/tex]Hence, the 5 arithmetic means between -4 and 8 are -2, 0, 2, 4, and 6, respectively.

Answered by GreatRedSpot | 2024-09-08