If four arithmetic means are inserted between -3 and 92, what is the second arithmetic mean?
Answer (1)
The second arithmetic mean in the given is 35.To find the answer, let's use the formula [tex]a_n=a_1+(n-1)d[/tex]Since there are 4 arithmetic means, there are 4 missing numbers. -2,_,_,_,_,92.We can see here that [tex]a_6[/tex] = 92. Let's use the equation. [tex]a_n=a_1+(n-1)d[/tex][tex]92 = -3+(6-1)d[/tex]Subtract the ones in the parenthesis and distribute d[tex]92 = -3 +5d[/tex]Transpose -3 to be positive 392+3 = 5d95 = 5dDivide both sides by 5[tex]\frac{95}{5} =\frac{5d}{5}[/tex]19=dNow that we have our d, we'll just add it twice to [tex]a_1[/tex]. another option would be to use the formula and substitute [tex]a_n[/tex] to [tex]a_3[/tex][tex]a_1[/tex] = -3[tex]a_2[/tex] = -3+19 = 16[tex]a_3[/tex]= 16+19 = 35Since the arithmetic mean is after [tex]a_1[/tex], to find the second arithmetic mean, it's [tex]a_3[/tex]. Therefore, the second arithmetic mean is 35.
