A) \[\frac{b-a}{n+1}\]
B) \[\frac{b+a}{n-1}\]
C) \[\frac{b-a}{n-1}\]
D) \[\frac{b+a}{n+1}\]
Correct Answer: A
Solution :
\[H.C.F.=\frac{\text{Product of the numbers}}{L.C.M.}\] \[\text{5474}=\text{2}\times \text{7}\times \text{17}\times \text{23}\] \[\text{9775}={{\text{5}}^{\text{2}}}\times \text{17}\times \text{23}\] \[\text{1173}0=\text{2}\times \text{3}\times \text{5}\times \text{17}\times \text{23}\] \[\therefore \] \[=\text{17}\times \text{23}=\text{391}\]
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