Question

Insert three arithmetic means between \[23\] and \[7\].

Answer 1

Hint: The arithmetic mean of any two numbers \[a\] and \[b\] is given by \[\dfrac{{a + b}}{2}\], such that the sequence \[a,A,b\] is in A.P. Here, in the given question, we have to insert three arithmetic means (AMs) in between \[23\] and \[7\] which implies the resultant Arithmetic Progression (A.P.) will be of total five terms. Now, at first, we will assume the three unknown values to form an A.P. and then we will use the formula of the general term of an A.P. to calculate the common difference \[d\]. As we have common differences now, we will use it to calculate the three required AMs.
Formula used:
Arithmetic Mean of any two numbers \[a\] and \[b\]= \[\dfrac{{a + b}}{2}\]
General term of an A.P. = \[a + \left( {n - 1} \right)d\]

Complete step-by-step solution:
Let \[{A_1},{A_2},{A_3}\] be the three numbers between \[23\] and \[7\] such that \[23,{A_1},{A_2},{A_3},7\] is an A.P.
Now, we know that,
General term of an A.P. = \[a + \left( {n - 1} \right)d\]
Then, fifth term of A.P.= \[a + 4d\]
In the above A.P., \[7\] is the fifth term, and \[a = 23\] is the first term, putting that value,
\[7 = 23 + 4d\]
Simplifying it, we get,
\[d = - 4\]
Now, using this\[d\]and general term formula, we will calculate the three required AMs,
General term of an A.P. = \[a + \left( {n - 1} \right)d\]
Second term of A.P. is calculated by \[a + d\]
\[ {A_1} = 23 + \left( { - 4} \right) \\
  {A_1} = 19 \]
Third term of the A.P. is calculated by, \[a + 2d\]
\[ {A_2} = 23 + 2\left( { - 4} \right) \\
  {A_2} = 15 \]
Fourth term of A.P. is calculated by, \[a + 3d\]
\[ {A_3} = 23 + 2\left( { - 4} \right) \\
  {A_3} = 11 \]
Hence, we can insert \[19,15,11\] between \[23\] and \[7\] so that the resulting sequence is in A.P.

Note: There is an interesting property of an arithmetic progression (A.P.) which states that if we add or subtract any constant number to/from each term of the series, we will still get a resultant series as A.P. with the same common difference.