Question

If A be an arithmetic mean between two numbers and S be the sum of n arithmetic means between the same numbers, then

• A. S - nA
• B. A = nS
• C. A = S
• D. None of these

Answer 1

The correct option is A S - nA

Let the two quantities be a and b and let $A_1,A_2.......,A_n$ be the n A.M.'s between them. Then $a,A_1,A_2.....A_n,b$ are in A.P. and let d be the common difference.
Now $T_{n+2}=b=a+(n+2-1)d\Rightarrow d=\frac{b-a}{n+1}$

Also $A_1+A_2+.....+A_n=S_{n+1}-a$
$=\frac{1}{2}(n+1)[2a+(n+1-1\left)\frac{(b-a)}{(n+1)}\right]-a$
$=\frac{n}{2}[2a+(b-a\left)\right]=\frac{n}{2}(a+b)=n\left(\frac{a+b}{2}\right)=nA.$
Trick: Let 1,3,5,7,9 is in A.P. In this series A = 5, n = 3, S = 15 $\Rightarrow$ S = nA.