Question

Find: \[3, 6, 21, 28, 55, 66, ?, 120\]
A) \[103\]
B) \[104\]
C) \[108\]
D) \[106\]
E) \[105\]

Answer 1

Hint:
This question can be easily solved by looking at the pattern of the two adjacent numbers. This can be done by finding their common factor, comparing it with that of the others and then looking for a meaning in the complete overall question.

Complete step by step solution:
These questions are solved by first observing the first k numbers of the sequence, that is to say look for a common factor in them and note it down. Then we do the same thing for the next k numbers till we have a nice and known sequence. What other important thing we got to do all along the way is maintain a new sequence which contains the first k numbers divided by their common factor. Then we separately compare the common factors and the new list containing the sequenced numbers and look for a pattern. Finally, we evaluate the answer by filling in the blank.

In this question, clearly, we can see that the k is \[2\]. So, the common factors change by two numbers.
The first two numbers are \[3\] and \[6\].
Clearly, the common factor for the two is \[3\].
We shall be maintaining a list for the common factors (say n list) and another list for the pair of two numbers divided by their common factor (say h list).
So, n = \[\left[ 3 \right]\]
And, h = \[\left[ {\dfrac{3}{3},\dfrac{6}{3}} \right] = \left[ {1,2} \right]\]
Then, the next two numbers are \[21\]and \[28\]. Their common factor is \[7\].
So, n = \[\left[ {3,7} \right]\]
And, h = \[\left[ {1,2,\dfrac{{21}}{7},\dfrac{{28}}{7}} \right] = \left[ {1,2,3,4} \right]\]
Then, the next two numbers are \[55\]and \[66\]. Their common factor is \[11\].
So, n = \[\left[ {3,7,11} \right]\]
And, h = \[\left[ {1,2,3,4,\dfrac{{55}}{{11}},\dfrac{{66}}{{11}}} \right] = \left[ {1,2,3,4,5,6} \right]\]

Clearly, the next number in the h list is \[15\](as the list follows the rule of hn+1 = hn + \[4\], and the one before \[15\] is \[11\]) and that in the n list is\[7\] (as they are just creating a list of consecutive natural number, and the one before \[7\]is \[6\]).
So, the required value is \[15 \times 7 = 105\].
Hence, the question mark in the question can be replaced by \[105\].

So, the correct option is E.

Note:
Such questions can be very easily solved by looking for a common factor of the first n numbers, and noting it down. Then, the next step is maintaining a new list of numbers which contain the first n numbers divided by their common factor. And finally, the last step is looking for a pattern or sequence in the common factors and the new sequence, separately, and filling in the blanks.