Latest TCS Aptitude Question SOLUTION: 53) The sequence {An } is defined by A1 = 2 and An+1 =An + 2n. What is the value of A100 ? a) 9902 b) 9900 c) 10100 d) 9904

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53) The sequence {An } is defined by A1 = 2 and An+1 =An + 2n. What is the value of A100 ?
a) 9902
b) 9900
c) 10100
d) 9904

Asked In TCS MAN (10 years ago)

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chandan kumar  (10 years ago)

ans is 9902
a1=2,An+1=An+2n
from this we can get series 2,4,8,14,22,32,....,
here 4-2=2,8-4=4,14-8=6,22-14=8,so difference is an a.p.
and 1st term + sum of differences=last term.
so use 2 as first term and 99 terms are there
sum of series of difference=(99/2)(2*2+(99-1)2)(A.P. formula to find sum of n terms)
hence A100=A1+sum of series of difference=2+9900=9902.

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Ann Mary Jolly  (10 years ago)

Ans : a)9902
From given,An = An-1 + 2(n-1)
since A1=2,An = 2 + 2*1 + 2*2 +2*3 +......2*(n-1)
Taking 2 common,An = 2*(1 + (1+2+3+......n-1) )
Now,1+2+3+.....n is given by n*(n+1)/2.
Therefore,1+2+3....n-1 = (n-1)*n/2
Therefore, An = 2*(1+n(n-1)/2) = 2+n(n-1) => An = 2+n(n-1)
Hence, A100 = 2+100*99 = 9902

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aditi  (10 years ago)

ans c)10100
see the sequence:
A(1)=2, A(2)=6, A(3)=12, A(4)=20... and so on..
the pattern we can see is
A(2)=2*(2+1)
A(3)=3*(3+1) and so on.. so
A(100)=100*(100+1)=101*100=10100

so ans is c

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Ashok kumar Gajula  (10 years ago)

please apply the concept of recursion:
A(n+1)=A(n)+2n
={A(n-1)+2(n-1)}+2n
={A(n-2}+2(n-2)}+2(n-1)+2n
=A(n-2)+2{(n-2)+(n-1)+n)}
...
....
=A(1)+2(1+2+.......+(n-1)+n)
=2+2(n(n+1)/2)
A(n+1)=2+n(n+1)
there4 A(100)=2+100(99)=9902.

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