A) \[ab\]
B) \[n(a+b)\]
C) \[nab\]
D) \[\frac{(a+b)}{n}\]
E) \[\frac{n(a+b)}{ab}\]
Correct Answer: B
Solution :
\[\because \] \[{{a}_{1}},{{a}_{2}},....,{{a}_{n}}\]are n AMs between a and b, then \[{{a}_{1}}=a+\frac{b-a}{n+1}=\frac{na+b}{n+1}\] and \[{{a}_{n}}=a+\frac{nb-na}{n+1}=\frac{nb+a}{n+1}\] \[\therefore \] \[2\sum\limits_{i=1}^{n}{{{a}_{i}}=2[{{a}_{1}}+{{a}_{2}}+....+{{a}_{n}}]}\] \[=2\left[ \frac{na+b}{n+1}+....+\frac{a+nb}{n+1} \right]\] \[=\frac{2}{n+1}\left[ \frac{n}{2}(na+b+a+nb) \right]\] \[=\left( \frac{n}{n+1} \right)[(n+1)(a+b)]\] \[=n(a+b)\]
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