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Question

Consider the sequence of numbers 2,5,5,8,8,8,11,11,11,11,.... The 150th term of the sequence is

A
48
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B
50
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C
47
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D
53
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Solution

The correct option is B 50
2 is coming one time
5 is coming two times
8 is coming four times and so on.
We can see that 2 is 1st term, 5 is 2nd term, 8 is 4th term, 11 is 7th term and so on.
Hence, (3k+2) is the (k(k+1)2+1)th term, where kI
Take k=16(k(k+1)2+1)=137k=17(k(k+1)2+1)=154

Hence, 150th term will be for k=16
The number will be 3k+2=3×16+2=50
Hence, 150th term is 50

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