A) 4, 5, 6
B) 3, 5, 7
C) 1, 5, 9
D) 2, 5, 8
Correct Answer: B
Solution :
Let three numbers are\[a-d,\ a,\ a+d\]. We get \[a-d+a+a+d=15\]\[\Rightarrow \]\[a=5\] and \[{{(a-d)}^{2}}+{{a}^{2}}+{{(a+d)}^{2}}=83\] \[\Rightarrow \] \[{{a}^{2}}+{{d}^{2}}-2ad+{{a}^{2}}+{{a}^{2}}+{{d}^{2}}+2ad=83\] \[\Rightarrow \]\[2({{a}^{2}}+{{d}^{2}})+{{a}^{2}}=83\] Putting \[a=5\] \[\Rightarrow \]\[2(25+{{d}^{2}})+25=83\]\[\Rightarrow \]\[2{{d}^{2}}=8\]\[\Rightarrow \]\[d=2\] Thus numbers are 3, 5, 7. Trick: Since \[3+5+7=15\] and \[{{3}^{2}}+{{5}^{2}}+{{7}^{2}}=83\].
You need to login to perform this action.
You will be redirected in
3 sec
