Nth term of a sequence = an + (n-1)d

So 36 will be the nth term of the sequence => 36 = 1+(n-1)k
where k is the common difference

=> 35 = (n-1)k
=>35/k=(-1)

So k will be a factor of 35 then only 36 be a possible term of the sequence

35 = 5x7 -----> that number of factors of 35= (1+1)(1+1)=4
therefore answer should be 4 .

Bunuel please check the answer options

chetan2u could you help solve this ?

sthahvi

The options do not contain the correct answer.

Now k is the common difference between consecutive terms.
Whatever be the difference between 1 and 36 should be equally distributable in k parts, then only k would be an integer.
36-1=35, so factors are 1,5,7 and 35.
The sequences are
1,2,3,4...35,36,37...
1,6,11,16....31,36,41...
1,8,15,22,29,36,43,...
1,36,71...

Since no options are anywhere close to 4, 36 must be some other value.

If I look at the options, 361 is one such value.
\(361-1=360=2^3*3^2*5\)
Number of factors = (3+1)(2+1)(1+1)=24

E
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The tough part is understanding the question layout and what the defined sequence really is.

The only answer it could possibly be is (E) because the max value for each Sk listed other than 24 is too low and less than 361.

(D)16

For S15, you will get 16 values.

The 1st value will be 1

Then the 2nd value will be: 1 + (1)(15) = 16

The 3rd value will be: 1 + (2)(15) = 31

And the 16th value will be: 1 + (15)(15) = 226

So the max number in the sequence is too low.

The answer must be E.

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