Latest TCS Aptitude Question SOLUTION: the sequence {An} is defined by A1=2 and A(n+1)=An+2n what is the value of A100........... 9902 9900 10100 9904

• Placement Papers
• /
• TCS

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40012047) TCS question Keep an EYE

the sequence {An} is defined by A1=2 and A(n+1)=An+2n
what is the value of A100...........

9902
9900
10100
9904

Asked In TCS Pawan (11 years ago)

Unsolved Read Solution (5)

Is this Puzzle helpful?   (5)   (1) Submit Your Solution

Read Solution (5) :

jayasri  (11 years ago)

a(n+1)-A(n)=2n
A(100)-A(99)=2*99
A(99)-A(98)=2*98
.
.
A(2)-A(1)=2*1
finally,by addind all of these,A(100)-A(1)=2*(99+98+....+1)
=2*(99*100/2)=9900
A(100)=9900+A(1)=9902

Is this solution Helpfull? Yes (34) | No   

hansa priya  (11 years ago)

A1=2 ,A(n+1)=An+2n
A2=2+2*1=4(i.e: 2^2-0)
A3=4+2*2=8(i.e: 3^2-1)
A4=8+2*3=14(i.e: 4^2-2)
A5=14+2*4=22(i.e: 5^2-3)
.
.

.
.
going so on we have A100=100^2-98=9902

Is this solution Helpfull? Yes (9) | No (1)  

shashank kansal  (11 years ago)

A(1)=2;
and formula is A(n+1)=A(n)+2*n;
now calculate A(2)it will be
A(2)=A(1+1)=A(1)+2*1=2+2*1=4
A(3)=A(2+1)=A(2)+2*2=4+2*2=8
A(4)=A(3+1)=A(3)+2*3=8+2*3=14
A(5)=A(4+1)=A(4)+2*4=14+2*4=22
hence the series will be
4,8,14,22
now the formula to generate this series will be..
4=(2^2)-0;
8=(3^2)-1;
14=(4^2)-2;
22=(5^2)-3;
i.e (n^2)-(n-2)
for 100
it will be
(100^100)-(100-2)
10000-98=9902...

Is this solution Helpfull? Yes (7) | No   

mahalakshmi  (11 years ago)

A1=2
A2=A(1+1)=An+2n=2+2=4 or 2^2-0
A3=4+4=8 or 3^2-1
A4=8+6=14 or 4^2-2
.
.
.
.
A100=100^2-98 = 9902

Is this solution Helpfull? Yes (6) | No   

Kishore Kumar  (11 years ago)

a(n+1)-A(n)=2n
A(100)-A(99)=2*99
A(99)-A(98)=2*98
.
.
A(2)-A(1)=2*1
finally,by addind all of these,A(100)-A(1)=2*(99+98+....+1)
=2*(99*100/2)=9900
A(100)=9900+A(1)=9902

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.