Latest Infosys Aptitude Question SOLUTION: Q. Find the 8th term in the series 3,11,31,69,131,....

• Placement Papers
• /
• Infosys
• /
• Numerical Ability
• /
• Sequence and Series

Infosys Question Papers: Infosys Interview Experience(10) INFOSYS Company Profile

(#M40031922) INFOSYS Numerical Ability question series Keep an EYE

Q. Find the 8th term in the series 3,11,31,69,131,....

Asked In Infosys Nitesh (10 years ago)

Solved prashanthi Read Solution (8)

Is this Puzzle helpful?   (29)   (16) Submit Your Solution Sequence and Series

Read Solution (8) :

prashanthi  (10 years ago)

1^3=1+2=3
2^3=8+3=11
3^3=27+4=31
4^3=64+5=69
5^3=125+6=131
6^3=216+7=223

Is this solution Helpfull? Yes (15) | No (1)  

SUGANDHA SAGAR  (10 years ago)

ans is 521

Is this solution Helpfull? Yes (7) | No (2)  

Akshay Bhandari  (10 years ago)

The answer is:521
Taking the double difference,
11-3=8
31-11=20
69-31=38
131-69=62
Now,Taking the double difference-
20-8=12
38-20=18
62-38=24
which is having a common difference of 6.

Is this solution Helpfull? Yes (5) | No (1)  

SUGANDHA SAGAR  (10 years ago)

3,11,31,69,131,.....
223

Is this solution Helpfull? Yes (1) | No (2)  

jyoti prajapati  (10 years ago)

1^3+2=3
2^3+3=11
3^3+4=31
4^3+5=69
5^3+6=131
6^3+7=223
7^3+8=351
8^3+9=521

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

subhasis jama  (10 years ago)

1056, as there are total 24 no which is divided by 4, U D places should be divisible by 4 for a 4 digit no.. we cant take 88 and 44 as rptn is not allowed.. so total 24-2=22 no of arrangement can be done for U-D places.. now for H T places it should be 6c1* 8c1... so total no of arrangements 8*6*22..= 1056

Is this solution Helpfull? Yes | No (5)  

karthikeyan  (10 years ago)

3,11,31,69,131,

8,20,38,62,92,128,

12,18,24,30,36,42,

ans-521

Is this solution Helpfull? Yes | No (2)  

Krishanu Bhattacharya  (10 years ago)

521
8^3+9=521

Is this solution Helpfull? Yes | No (1)  

Please Login to Read Solution.