Question

Given two numbers a and b, let A denotes the single A.M. and S denotes the sum of n A.M.s between a and b, then $\dfrac{S}{A}$ depends on
A. \[n,a,b\]
B. \[n,b\]
C. \[n,a\]
D. $n$

Answer 1

Hint: First find the arithmetic mean of the two numbers. After that for finding the sum of n number of arithmetic mean between a and b use the summation formula $S = \dfrac{n}{2}\left( {{a_1} + {a_n}} \right)$. Substitute the values and get the sum. After that substitute, the values in $\dfrac{S}{A}$ and do simplification to get the desired result.

Formula used: The formula for arithmetic mean is,
$A = \dfrac{{x + y}}{2}$
Where x, y is the numbers.
The formula for the sum of the arithmetic term is,
$S = \dfrac{n}{2}\left( {{a_1} + {a_n}} \right)$
Where ${a_1}$ is the first term
${a_n}$ is the last term
n is the number of terms

Complete step-by-step solution:
As it is given that A is the single A.M. between a and b. Then,
$ \Rightarrow A = \dfrac{{a + b}}{2}$
Now, let there is n A.M.s between a and b and the common difference be d,
$ \Rightarrow a,{A_1},{A_2},{A_3}, \ldots ,{A_n},b$
Now evaluate the sum of terms from ${A_1}$ to ${A_n}$. Then,
$ \Rightarrow S = \dfrac{n}{2}\left( {{A_1} + {A_n}} \right)$.............….. (1)
As the A.M. are in arithmetic progression. So,
$ \Rightarrow {A_1} = a + d$
Also,
$ \Rightarrow b = {A_n} + d$
Move d on the other side,
$ \Rightarrow {A_n} = b - d$
Substitute the values in the equation (1),
$ \Rightarrow S = \dfrac{n}{2}\left( {a + d + b - d} \right)$
Simplify the terms,
$ \Rightarrow S = \dfrac{n}{2}\left( {a + b} \right)$
Now, we have to find $\dfrac{S}{A}$. So, substitute the values,
$ \Rightarrow \dfrac{S}{A} = \dfrac{{\dfrac{n}{2}\left( {a + b} \right)}}{{\dfrac{{a + b}}{2}}}$
Cancel out the common terms,
$\therefore \dfrac{S}{A} = n$
Thus, $\dfrac{S}{A}$ depends on $n$ only.

Hence, option (D) is the correct answer.

Note: While solving this question, the possible mistakes we can make is in the sum formula, that is, we might think of using $S = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$. But we do not have the value of d. So, we will get stuck somewhere. So, we will prefer to use $S = \dfrac{n}{2}\left( {a + l} \right)$, where l is the last term. So, we have to be careful while solving the question.