sequence

[demoskopico]

In an infinite sequence of integers a1, a2, a3,… , a1 = -30 and for all n > 1, an-1 + 4 <an < an-1 +8. If ak = 47, how many possibilities are there for the value of k?

(A) Four (B) Five (C) Six (D) Seven (E) Eight


how can I find the quickest way? is it very tough or I am too weak?

[kevin]

Do you agree that each term is either 5 ,6 or 7 greater than the preceding term?

Therefore, 47, the kth term is an integer from 5(k-1) -30 to 7(k-1) - 30.

This means that 5(k-1) =< 77 and 7(k-1) >= 77

Can you continue?

[demoskopico]

Do you agree that each term is either 5 ,6 or 7 greater than the preceding term?

Therefore, 47, the kth term is an integer from 5(k-1) -30 to 7(k-1) - 30.

This means that 5(k-1) =< 77 and 7(k-1) >= 77

Can you continue?

I agree that each term is 5,6 or 7 greater, but this would mean that 47 could be a(k-1) + 7 at his maximum level and a(k-1) + 3 at his low. how come that 47, the kth term is an integer from 5(k-1) -30 to 7(k-1) - 30?

[kevin]

Suppose each term were 5 greater than the previous term. What would be the kth term?

[demoskopico]

Suppose each term were 5 greater than the previous term. What would be the kth term?


it would be (k-1) + 5, then?


what do you thin about this reasoning?

ak - a(k-1) could be a value included between 7 and 5
let's go with the difference of 7. we know that ak - a1=77. since the difference at limit is 7, we must divide 77 by 7 in order to find k, thus k=11

same reasoning for 5, thus we have 15 as top limit

values can be 11,12,13,14,15....is that so bad?