Question

The sum of \[n\] arithmetic means between \[a\] and \[b\], is
1) \[\dfrac{{n\left( {a + b} \right)}}{2}\]
2) \[n\left( {a + b} \right)\]
3) \[\dfrac{{\left( {n + 1} \right)\left( {a + b} \right)}}{2}\]
4) \[\left( {n + 1} \right)\left( {a + b} \right)\]

Answer 1

Hint: Here, we will calculate the sum of arithmetic means using the formula of finding the sum of arithmetic means of the \[n + 2\] numbers is \[\left( {\dfrac{{n + 2}}{2}} \right)\left( {a + b} \right)\]. Then we will subtract the obtained sum by \[a + b\] on each side to find the required solution. Apply these formulas, and then use the given conditions to find the required value.

Complete step-by-step answer:

Let us assume that the two numbers be \[a\] and \[b\], the arithmetic means are \[{A_1}\], \[{A_2}\], \[{A_3}\], …, \[{A_n}\] between \[a\] and \[b\].

Then, we will have the arithmetic means \[a\], \[{A_1}\], \[{A_2}\], \[{A_3}\], …, \[{A_n}\], \[b\], which are in arithmetic progression A.P.

We know that \[a\] be the first term and \[b\] be the second term in the given arithmetic progression.

Since there are \[n + 2\] numbers in the arithmetic progression, so we know that the formula to find the sum of arithmetic means of \[n + 2\] numbers is \[\left( {\dfrac{{n + 2}}{2}} \right)\left( {a + b} \right)\], where \[a\] is the first term and \[b\] is the last term.

Using this formula to find the arithmetic mean, we get

\[a + {A_1} + {A_2} + {A_3} + ... + {A_n} + b = \left( {\dfrac{{n + 2}}{2}} \right)\left( {a + b} \right)\]

Subtracting the above equation by \[a + b\] on each of the sides to find the sum of \[n\] arithmetic means.

\[
   \Rightarrow a + {A_1} + {A_2} + {A_3} + ... + {A_n} + b - \left( {a + b} \right) = \left( {\dfrac{{n + 2}}{2}} \right)\left( {a + b} \right) - \left( {a + b} \right) \\
   \Rightarrow a + {A_1} + {A_2} + {A_3} + ... + {A_n} + b - a - b = \left( {a + b} \right)\left[ {\left( {\dfrac{{n + 2}}{2}} \right) - 1} \right] \\
   \Rightarrow {A_1} + {A_2} + {A_3} + ... + {A_n} = \left( {a + b} \right)\left[ {\dfrac{{n + 2 - 2}}{2}} \right] \\
   \Rightarrow {A_1} + {A_2} + {A_3} + ... + {A_n} = \left( {a + b} \right)\left[ {\dfrac{n}{2}} \right] \\
   \Rightarrow {A_1} + {A_2} + {A_3} + ... + {A_n} = n\left( {\dfrac{{a + b}}{2}} \right) \\
\]

Thus, the sum of \[n\] arithmetic means between \[a\] and \[b\] is \[n\left( {\dfrac{{a + b}}{2}} \right)\].

Hence, the option is A is correct.

Note: In solving these types of questions, student use the formula to find the sum, \[S = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\], but we do not have the value of \[d\]. So, we will here use \[S = \dfrac{n}{2}\left( {a + l} \right)\], where \[d\] is the last term. Also while calculating the sum, we might make a mistake by writing the number of terms as \[n\] instead of \[n + 1\]. So, we have to be careful while solving the question.