Question
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 AS depends on
A. n,a,b
B. n,b
C. n,a
D. n
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=2n(a1+an). Substitute the values and get the sum. After that substitute, the values in AS and do simplification to get the desired result.
Formula used: The formula for arithmetic mean is,
A=2x+y
Where x, y is the numbers.
The formula for the sum of the arithmetic term is,
S=2n(a1+an)
Where a1 is the first term
an 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=2a+b
Now, let there is n A.M.s between a and b and the common difference be d,
⇒a,A1,A2,A3,…,An,b
Now evaluate the sum of terms from A1 to An. Then,
⇒S=2n(A1+An).............….. (1)
As the A.M. are in arithmetic progression. So,
⇒A1=a+d
Also,
⇒b=An+d
Move d on the other side,
⇒An=b−d
Substitute the values in the equation (1),
⇒S=2n(a+d+b−d)
Simplify the terms,
⇒S=2n(a+b)
Now, we have to find AS. So, substitute the values,
⇒AS=2a+b2n(a+b)
Cancel out the common terms,
∴AS=n
Thus, AS 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=2n[2a+(n−1)d]. But we do not have the value of d. So, we will get stuck somewhere. So, we will prefer to use S=2n(a+l), where l is the last term. So, we have to be careful while solving the question.