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
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5 comments:
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
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
its 802....
first term is 2....
second is 6.....
thrid is 10, 14, 18.....
so gap is of 4.
so 201*4 = 802
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..
802.
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