Question

Fool's problem of the day,

Today's problems are to find the next term(s) of some given series

(a) \( 7,9,40,74,1526, ? \)

\( 5436 \quad\quad 6486 \quad\quad5486 \quad\quad3456 \)

(b) \( 1,3,8,18,? \)

\( 26 \quad\quad 30 \quad\quad 32 \quad\quad 38 \)

(c) \( Z, X, S,I,R,R,?,? \)

\( JK \quad\quad JI \quad\quad KM \quad\quad GI \)

Please post the solution instead of just the answer.

PS: Only the first one is relatively easy/well-known.

Answer 1

fool never fails to disappoint..always nearly unsolveable =)))

Answer 2

for (b), 35?

Answer 3

@CoCoTsoi: Sorry, 35 is the not the right answer.

Answer 4

Happy to learn from failure :D

Answer 5

these are not like college exams that have common patterns...the progressions seems inconsistent..but still seems not...

Answer 6

it is quite hard to know whether we think it correctly coz the sequence given can deduct numerous sequence that are different

Answer 7

for (b), 68?

Answer 8

@CoCoTsoi: I have added option now :)

Answer 9

that's excellent!

Answer 10

(b) 26

Answer 11

Please post solution instead of just an answer.

Answer 12

go Coco! be the first person to finally answer Fool's question for the day! represent OpenStudy!!! hahaha =)))

Answer 13

for b) tn = tn+(n^2+1) works out ... but there not in the answer

Answer 14

i dont know what these mean..goodluck with these =)))) <insert yaoming meme>

Answer 15

Help, I am confused by myself. Wait a little bit longer plz

Answer 16

tn = t(n-1)+((n-1)^2+1)

Answer 17

No one has posted the correct answer to any one of them.

Answer 18

hark working very very working ...!!

Answer 19

and you're wondering....why? lol =))) hey fool..has anyone answered one of your Fool's problem of the day?

Answer 20

@lgbasallote: Yes many did.

Answer 21

man .. i was wondering who what order would give such plain and sharp rise

Answer 22

wow..they must feel proud...lemme guess...ishaan, diyadiya, saifoo, bahrom, those folks right?

Answer 23

For a, is it 5436?
I dunno what i'm doing..
third term = (1st term)^2 - second term = 7^2 -9
4th term = 2nd term ^2 - first term
5th term = 3rd term^2 - 4th term
So i'm guessing 6th term = 4th term^2 - 3rd term
@Callisto, it's not correct :(

Answer 24

why did callisto tag herself? i wonder....hahaha

Answer 25

for b) 32?

Answer 26

I would have chosen 35 for b as well...

Answer 27

i think, a) 5436

Answer 28

For the first one, I thought the same way Callisto did.
For the second one, I guess 26(The terms are 1^2-0,2^2-1,3^2-1,4^2+2, 5^2+1)
@order and @anonymoustwo44, Please post your explanations.

Answer 29

@Mani_Jha but then that's doesn't look like a series for b

Answer 30

*delete ' 's ' please*

Answer 31

I thought (b) 1,3,8,18,35

since, there's a difference of 2, 5, 10, 17, since
1-2 = 1, 2-5=3, 5-10=5, so, 10-x = 7 = 17,

Answer 32

I know. That's why I said I 'guess' :P

Answer 33

callisto is right for the first one! Congratz :D

@anonymoustwo44: Solution please.

Answer 34

Why isn't 42 in the options? hehe

Answer 35

i wish there could be a 31 in the option :(

Answer 36

maybe this,\[a_0 = 1\]
\[a_{n} = n^2 + 1 + a_{n-1}, n \in \mathbb{N}\]

Answer 37

but 35 isn't the option

Answer 38

I hate these questions

Answer 39

for b)
2^0=1, 1+2^1=3, 3+2^2+2^0=6,8+2^1+2^3=18. 18+2^2+2^4 = 38?

Answer 40

correction :
2^0=1, 1+2^1=3, 3+2^2+2^0=8,8+2^1+2^3=18, 18+2^2+2^4 = 38?

Answer 41

Can you generalize your series?

Answer 42

I seeing the answer to b) as 30 but am having difficulty explaining the first two terms

Answer 43

1*3=3
3*3-1*1=8
8*3-2*3=18
18*3-3*8=30

Answer 44

I guess some weird formula like a(n)=a(n-1)+2^(n-3)+2^(n-1), but I don't know if I can just not take into account the negative exponants :-S

Answer 45

b)
1=1
1*3=3
3*3-1*1=8
8*3-2*3=18
18*3-3*8=30\[x_0=1\]\[x_n=3x_{n-1}-(n-1)x_{n-2}\]maybe, looks too simple though

Answer 46

For 3rd it's GI

Answer 47

how do you figure?

Answer 48

I think I'm right for b) hehe

Answer 49

It's relatively easy, just list out the alphabets, series follow \(a_n = n^2 + 1, n \in \mathbb{N}\) from the bottom.

Answer 50

or, Z to A.

Answer 51

I see :)

Answer 52

a(n+1) = a(n-1)^2 - a(n - 2*(1 - n mod 2)), a(0)=7, a(1)=9

Answer 53

Congratz! ishaan, you got the right answer of (3)

Answer 54

For 2: Answer is 32 and hint is "Prime numbers"

Answer 55

aww... I thought I had it for once!
I do realize I had a problem with the fact that \(x_{-1}\) is not defined
ok I'll keep staring at it...

Answer 56

3+5 = 8, 7+11 = 18, 13+19 = 32, so it looks a lot like it's ((n-1)*2-2)th + ((n-1)*2-1)th primes, but that doesn't really explain 1 and 3, does it?

Answer 57

Are these polynomials?

Answer 58

@inkyvoyd: Interpolation won't help.

Answer 59

*I'm out*

Answer 60

I give up :/ ...even after hint I wasn't able to do it :( ...Post the logic FoolForMath if others think so too, I know I am gonna feel stupid after you post the solution but feeling stupid for an hour is better than feeling hopeless for several hours.

Answer 61

1+2=3
3+5=8
7+11=18
13+17=30
with that way of summing pairs of consecutive primes I got 30 again lol
it looks like you want me to sum 19 and 13, but why...?

Answer 62

I agree with Ishaan, I'm really curious about this

Answer 63

Solution of (b):

Here, in this series we are omitting prime numbers in an increasing order. Between 1 and 3 we are omitting '2' i.e., one prime number. Between 3 and 8 we are omitting 5 and 7 i.e. two prime numbers. Similarly between 18 and missing number we have to omit 4 prime numbers. The next four prime numbers after 18 are 19,23,29,31. Hence,32 is the missing term.

Answer 64

That's it? A very unusual series.

Answer 65

I agree:)