A) \[\frac{a(n+1)}{2}\]
B) \[\frac{a(n-1)}{2}\]
C) \[\frac{an}{2}\]
D) None of these
Correct Answer: C
Solution :
[c] \[\frac{A{{L}_{1}}}{AB}=\frac{{{L}_{1}}{{M}_{1}}}{BC}\] |
\[\therefore \frac{1}{n+1}=\frac{{{L}_{1}}{{M}_{1}}}{a}\] |
\[\therefore {{L}_{1}}{{M}_{1}}=\frac{a}{n+1};\] |
\[\frac{A{{L}_{2}}}{AB}=\frac{{{L}_{2}}{{M}_{2}}}{BC}\] |
\[\therefore \frac{2}{n+1}=\frac{{{L}_{2}}{{M}_{2}}}{a}\] |
\[\therefore {{L}_{2}}{{M}_{2}}=\frac{2a}{n+1},\,\]etc. |
\[\therefore \] The required sum\[=\frac{a}{n+1}+\frac{2a}{n+1}+\frac{3a}{n+1}+\]\[...+\frac{na}{n+1}\] |
\[=\frac{a}{n+1}(1+2+....+n)=\frac{a}{n+1}.\frac{n(n+1)}{2}=\frac{an}{2}\] |
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