10 | 11 | 19 | ? | 92 | 117 |
A) 26
B) 28
C) 21
D) 29
E) 31
Correct Answer: B
Solution :
The series is \[+{{1}^{2}},+{{2}^{3}},+{{3}^{2}},+{{4}^{3}},+{{5}^{2}},+{{6}^{3}},.....\] ie \[\begin{align} & 10+{{1}^{2}}=11,11+{{2}^{3}}=19,19+{{3}^{2}}=\mathbf{28}, \\ & 28+{{4}^{3}}=92,92+{{5}^{2}}=117,...... \\ \end{align}\]You need to login to perform this action.
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