A) \[S=n\,A\]
B) \[A=n\,S\]
C) \[A=S\]
D) None of these
Correct Answer: A
Solution :
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 \[{{A}_{1}}+{{A}_{2}}+......+{{A}_{n}}={{S}_{n+1}}-a\] \[=\frac{1}{2}(n+1)\left[ 2a+(n+1-1)\frac{(b-a)}{(n+1)} \right]-a\] =\[\frac{n}{2}[2a+(b-a)]=\frac{n}{2}(a+b)=n\left( \frac{a+b}{2} \right)=nA\]. Trick: Let 1, 3, 5, 7, 9 is in A.P. In this series \[A=5,n=3,S=15\] Þ \[S=nA\].
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