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Find the fourth arithmetic means between -21 and -36

\(\text{We use $a_n = a_1 + (n-1)d$ to find the common difference $d$.} \)

Solution

\(\text{The first term is $a_1 = - 21$ and the fourth term is $a_4 = - 36 $.}\\ \text{We must find the common difference so that the terms }\)

\(\begin{array}{cccc} -21, & -21+d, & -21+2d, & -36 \\ \uparrow & \uparrow & \uparrow & \uparrow \\ a_1 & a_2 & a_3 & a_4 \end{array}\)

\(\text{form an arithmetic sequence.} \)

\(\text{To find the common difference $d$, we substitue $-21$ for $a_1$; 4 for $n$, and $-36$ for $a_n \\$in the formula for the 4th term:}\)

\(\begin{array}{rcll} a_4 &=& a_1 + (n-1)d \qquad & \text{This gives the 4th term of any arithmetic sequence.} \\ -36 &=& -21 + (4-1)d \qquad & \text{Substitute.} \\ -36 &=& -21 + 3d \qquad & \text{Subtract within the parentheses.}\\ -15 &=& 3d \qquad & \text{Subtract -21 from both sides.} \\ -5 &=& d \qquad & \text{To isolate d, divide both sides by 3.} \\ \end{array}\)

\(\text{To find the two arithmetic means between $-21$ and $-36$, $\\$we add the common difference $-5$, as shown:}\)

\(\begin{array}{rcll} -21+d &=& -21 +(-5) \\ &=& -26 \qquad & \text{This is $a_2$.} \\ \end{array}\)

\(\begin{array}{rcll} -21+2d &=& -21 +2(-5) \\ &=& -21 -10 \\ &=& -31 \qquad & \text{This is $a_3$.} \\ \end{array}\)

\(\text{Two arithmetic means between $-21$ and $-36$ are $-26$ and $-31$.} \)