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HARISH R

· started a discussion

· 1 Months ago

Sequence issue (Question: 99234.0 )

IF all are given in sequence then it can be assumed to solve the next in sequence and not in between.If solution is required then the question mark should have been placed at the right place in the sequence than by merely asking to solve.

Question:
Solve the series given below.


\(\cfrac{4}{3}, \cfrac{9}{15}, \cfrac{14}{27}, \cfrac{24}{51}\)
Options:
A) \(\cfrac{15}{39}\)
B) \(\cfrac{19}{39}\)
C) \(\cfrac{16}{37}\)
D) \(\cfrac{18}{37}\)
E) None of these
Solution:
(b) The series is : 

\(\cfrac{4+5}{3\times5}=\cfrac{9}{15}\)

\(\cfrac{9+5}{3\times9}=\cfrac{14}{27}\)

\(\cfrac{14+5}{3\times13}=\cfrac{19}{39}\)

\(\cfrac{19+5}{3\times17}=\cfrac{24}{51}\)

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