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.