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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 ...

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 SA\dfrac{S}{A} depends on
A. n,a,bn,a,b
B. n,bn,b
C. n,an,a
D. nn

Explanation

Solution

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=n2(a1+an)S = \dfrac{n}{2}\left( {{a_1} + {a_n}} \right). Substitute the values and get the sum. After that substitute, the values in SA\dfrac{S}{A} and do simplification to get the desired result.

Formula used: The formula for arithmetic mean is,
A=x+y2A = \dfrac{{x + y}}{2}
Where x, y is the numbers.
The formula for the sum of the arithmetic term is,
S=n2(a1+an)S = \dfrac{n}{2}\left( {{a_1} + {a_n}} \right)
Where a1{a_1} is the first term
an{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,
A=a+b2\Rightarrow A = \dfrac{{a + b}}{2}
Now, let there is n A.M.s between a and b and the common difference be d,
a,A1,A2,A3,,An,b\Rightarrow a,{A_1},{A_2},{A_3}, \ldots ,{A_n},b
Now evaluate the sum of terms from A1{A_1} to An{A_n}. Then,
S=n2(A1+An)\Rightarrow S = \dfrac{n}{2}\left( {{A_1} + {A_n}} \right).............….. (1)
As the A.M. are in arithmetic progression. So,
A1=a+d\Rightarrow {A_1} = a + d
Also,
b=An+d\Rightarrow b = {A_n} + d
Move d on the other side,
An=bd\Rightarrow {A_n} = b - d
Substitute the values in the equation (1),
S=n2(a+d+bd)\Rightarrow S = \dfrac{n}{2}\left( {a + d + b - d} \right)
Simplify the terms,
S=n2(a+b)\Rightarrow S = \dfrac{n}{2}\left( {a + b} \right)
Now, we have to find SA\dfrac{S}{A}. So, substitute the values,
SA=n2(a+b)a+b2\Rightarrow \dfrac{S}{A} = \dfrac{{\dfrac{n}{2}\left( {a + b} \right)}}{{\dfrac{{a + b}}{2}}}
Cancel out the common terms,
SA=n\therefore \dfrac{S}{A} = n
Thus, SA\dfrac{S}{A} depends on nn 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=n2[2a+(n1)d]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=n2(a+l)S = \dfrac{n}{2}\left( {a + l} \right), where l is the last term. So, we have to be careful while solving the question.