A) 190
B) 262
C) 225
D) 157
Correct Answer: A
Solution :
\[\frac{n(n+1)}{2}+99={{(n-2)}^{2}}\] \[{{n}^{2}}+n+198=2({{n}^{2}}+4-4n)\] \[{{n}^{2}}-9n-190=0\] \[{{n}^{2}}-19n+10-190=0\] \[n(n-19)+10(n-19)=0\] \[n=19\]You need to login to perform this action.
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