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

ans is 521
10 years agoHelpfull: Yes(7) No(2)
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.
10 years agoHelpfull: Yes(5) No(1)
3,11,31,69,131,.....
223
10 years agoHelpfull: Yes(1) No(2)
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
10 years agoHelpfull: Yes(1) No(1)
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
10 years agoHelpfull: Yes(0) No(5)
3,11,31,69,131,

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

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

ans-521
10 years agoHelpfull: Yes(0) No(2)
521
8^3+9=521
10 years agoHelpfull: Yes(0) No(1)

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