Wednesday, March 14, 2007

"QUESTION ON DEMAND"

What is the value 201st term of a sequence if
the first term of the sequence is 2 and each
successive term is 4 more the term immediately
preceding it?
(A) 798
(B) 800
(C) 802
(D) 804
(E) 806


contributed by DIVYA TOMAR

5 comments:

  1. i think the answer is 802 coz the numers follow a series with unit digit varyin from 2 6 0 4 8 and again the sme pattern so the 200 term will have unit digit8 and 201 will have 2 so only 802 is right

    ReplyDelete
  2. it is an arithmetic progreesion and the formula is Tn = a + (n-1)d
    where a = first term = 2
    ; Tn the nth term i.e. 201st term and
    d= common difference = 4
    therefore Tn = 2 + 200*4
    = 802

    ReplyDelete
  3. its 802....
    first term is 2....
    second is 6.....
    thrid is 10, 14, 18.....

    so gap is of 4.

    so 201*4 = 802

    ReplyDelete
  4. the ans is C=802

    note the formula guyz

    N=a+(n-1)*d

    where,N=req no
    a=1st no in seq
    d=common difference
    nth term..

    ReplyDelete