B. SOLVE THE FOLLOWING:1. INSERT 5 ARITHMETIC MEANSBETWEEN-4 AND 38.-4....382. INSERT 4 ARITHMETIC MEANSBETWEEN-12 AND 23.-12,..23​

1. SOLUTION:Step 1: List the given values.Since we must insert 5 arithmetic means between -4 and 38, the number of terms (5 arithmetic means + 2 given terms) is 7.[tex]\begin{aligned} & a_1 = -4 \\ & a_7 = 38 \\ & 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 \\ 38 & = -4 + 6d \\ -4 + 6d & = 38 \\ 6d & = 38 + 4 \\ 6d & = 42 \\ \frac{6d}{6} & = \frac{42}{6} \\ d & = 7 \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 + 7 \\ & = \boxed{3} \end{aligned}[/tex]• For a₃[tex]\begin{aligned} a_3 & = a_1 + (3 - 1)d \\ a_3 & = a_1 + 2d \\ a_3 & = -4 + 2(7) \\ a_3 & = -4 + 14 \\ & = \boxed{10} \end{aligned}[/tex]• For a₄[tex]\begin{aligned} a_4 & = a_1 + (4 - 1)d \\ a_4 & = a_1 + 3d \\ a_4 & = -4 + 3(7) \\ a_4 & = -4 + 21 \\ & = \boxed{17} \end{aligned}[/tex]• For a₅[tex]\begin{aligned} a_5 & = a_1 + (5 - 1)d \\ a_5 & = a_1 + 4d \\ a_5 & = -4 + 4(7) \\ a_5 & = -4 + 28 \\ & = \boxed{24} \end{aligned}[/tex]• For a₆[tex]\begin{aligned} a_6 & = a_1 + (6 - 1)d \\ a_6 & = a_1 + 5d \\ a_6 & = -4 + 5(7) \\ a_6 & = -4 + 35 \\ & = \boxed{31} \end{aligned}[/tex]Hence, the 5 arithmetic means between -4 and 38 are 3, 10, 17, 24, and 31, respectively.-----------------------------------------------------2. SOLUTION:Step 1: List the given values.Since we must insert 4 arithmetic means between -12 and 23, the number of terms (4 arithmetic means + 2 given terms) is 6.[tex]\begin{aligned} & a_1 = -12 \\ & a_6 = 23 \\ & n = 6 \end{aligned}[/tex]Step 2: Determine the common difference.[tex]\begin{aligned} a_n & = a_1 + (n - 1)d \\ a_6 & = -12 + (6 - 1)d \\ 23 & = -12 + 5d \\ -12 + 5d & = 23 \\ 5d & = 23 + 12 \\ 5d & = 35 \\ \frac{5d}{5} & = \frac{35}{5} \\ d & = 7 \end{aligned}[/tex]Step 3: Solve for the 4 arithmetic means.• For a₂[tex]\begin{aligned} a_2 & = a_1 + (2 - 1)d \\ a_2 & = a_1 + d \\ a_2 & = -12 + 7 \\ & = \boxed{-5} \end{aligned}[/tex]• For a₃[tex]\begin{aligned} a_3 & = a_1 + (3 - 1)d \\ a_3 & = a_1 + 2d \\ a_3 & = -12 + 2(7) \\ a_3 & = -12 + 14 \\ & = \boxed{2} \end{aligned}[/tex]• For a₄[tex]\begin{aligned} a_4 & = a_1 + (4 - 1)d \\ a_4 & = a_1 + 3d \\ a_4 & = -12 + 3(7) \\ a_4 & = -12 + 21 \\ & = \boxed{9} \end{aligned}[/tex]• For a₅[tex]\begin{aligned} a_5 & = a_1 + (5 - 1)d \\ a_5 & = a_1 + 4d \\ a_5 & = -12 + 4(7) \\ a_5 & = -12 + 28 \\ & = \boxed{16} \end{aligned}[/tex]Hence, the 4 arithmetic means between -12 and 23 are -5, 2, 9, and 16, respectively.

Answered by GreatRedSpot | 2024-09-01