Question

Find the $ 3 $ Arithmetic Means between $ 18\,\,and\,\,30 $ .

Answer 1

Hint: In this we first let three arithmetic means as $ {A_1},{A_{2\,\,}}and\,\,{A_3} $ . Then writing all three arithmetic means between given numbers to form an arithmetic progression and then solving value of common difference (d) using given value of first term (a) and hence using both value of ‘a’ and ‘d’ to get value of $ {A_1},{A_{2\,\,}}and\,\,{A_3} $ and so all three arithmetic means or we can say solution of given problem.
Formulas used: nth term in arithmetic progression: $ {T_n} = a + \left( {n - 1} \right)d, $ where ‘a’ is the first term of the series and ‘d’ is common difference of series.

Complete step-by-step answer:
Given numbers are $ 18\,\,and\,\,30 $ .
Let three arithmetic means between two given numbers or we can say between $ 18\,\,and\,\,30 $ are $ {A_1},{A_2}\,\,and\,\,{A_3} $ .
Therefore, we have an A.P. series. Which can be written as:
 $ 18,{A_1},{A_2},{A_3},30 $
Clearly we see that there are five terms in the above arithmetic progression.
Therefore we can say that $ {T_5} $ or fifth term of the series is $ 30 $
Also, we can see that first term of the series is $ 18. $
Let d be the common difference of given arithmetic progression.
Then from above or from the given fifth term. We have,
 $
  {T_5} = 30 \\
  or\,we\,\,can\,write\,\,it\,\,as: \\
  a + 4d = 30 \;
  $
Substituting value of ‘a’ in above equation.
 $
  18 + 4d = 30 \\
   \Rightarrow 4d = 30 - 18 \\
   \Rightarrow 4d = 12 \\
   \Rightarrow d = \dfrac{{12}}{4} \\
   \Rightarrow d = 3 \;
  $
Hence, from above we see that the common difference of given or formed A.P series is $ 3. $
Also, we see that the first arithmetic mean is the second term of an A.P.
Therefore we can write the first arithmetic mean as the second term. We have
 $
  {A_1} = {T_2} \\
   \Rightarrow {A_1} = a + d \;
  $
Now, substituting the value of ‘a’ and ‘d’ in the above equation.
 $
  {A_1} = 18 + 3 \\
   \Rightarrow {A_1} = 21 \;
  $
Also, the second arithmetic mean is the third term.
Therefore, we have
 $
  {A_2} = {T_3} \\
   \Rightarrow {A_2} = a + 2d \;
  $
Substituting value of ‘a’ and ‘d’ in above equation.
 $
  {A_2} = 18 + 2(3) \\
   \Rightarrow {A_2} = 18 + 6 \\
   \Rightarrow {A_2} = 24 \;
  $
Also, the third arithmetic mean is the fourth term of A.P.
Therefore, we have
 $
  {A_3} = {T_4} \\
   \Rightarrow {A_3} = a + 3d \;
  $
Substituting value of ‘a’ and ‘d’ in above we have,
 $
  {A_3} = 18 + 3(3) \\
   \Rightarrow {A_3} = 18 + 9 \\
   \Rightarrow {A_3} = 27 \;
  $
Therefore, from above we see that three arithmetic means between given numbers are $ 21,\,\,24\,\,and\,\,\,27 $ .
So, the correct answer is “ $ 21,\,\,24\,\,and\,\,\,27 $ .”.

Note: We can also find a solution to a given problem or we can say three arithmetic means between given numbers in other ways. This is a definition method. In this method we use a formula of arithmetic mean between two numbers. This is given as $ \dfrac{{a + b}}{2} $ . With the help of this formulas we first find arithmetic mean between $ 18\,\,and\,\,30 $ and then we use again same formulas between number $ 18 $ and number calculated earlier to get an arithmetic between two numbers and also we do same for $ 30 $ and number obtained in first step. Hence, we will get three arithmetic means between given numbers and solution of given problem.