| Directions: In each of these questions, a number series is given. In each series, only one number is wrong. Find out the wrong number. [SBI (PO) 2011] |
| 3601, 3602, 1803, 604, 154,36,12 |
A) 3602
B) 1803
C) 604
D) 154
E) 36
Correct Answer: D
Solution :
| \[[3601+{{(1)}^{2}}]\times 1=3602\] |
| \[\Rightarrow \]\[[3602+{{(2)}^{2}}\times \frac{1}{2}=1803\]\[\Rightarrow \]\[[1803+{{(3)}^{2}}\times \frac{1}{3}=604\] |
| \[\Rightarrow \]\[[604+{{(4)}^{2}}+\frac{1}{4}=155\ne 154\] |
| \[\Rightarrow \]\[[155+{{(5)}^{2}}]\times \frac{1}{5}=36\]\[\Rightarrow \]\[[36+{{(6)}^{2}}\times \frac{1}{6}=12\] |
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