GMAT Club World Cup 2022 (DAY 5): The infinite sequence of integers

I can't get the correct one. Guess C and move on

Given A1=-17
then putting the value of N in Inequality Equation for maximum value of N :
-24<A2<-19
-31<A3<-21
-38<A4<-23
-45<A4>-25
-52<A4>-27
-59<A4>-29
-66<A5<-31
similarly for 7 Values of N the Equation shall satisfy
Hence it will
Hence Only one value

\(a_1 = -17\)
\(a_x = -53\)

Using the AP formula,
\(a_x = a_1 + (x-1)*d\)
\((x-1)*d = -53+17\)
\((x-1)*d = -36\)

\(a_{n-1} - 7 < a_n < a_{n-1} - 2\)
Hence, \(d = {-6, -5, -4, -3}\)

\(x_{min} -> (x_{min} - 1) * -6 = -36\)
\(x_{min} = 6+1 = 7\)

\(x_{max} -> (x_{max} - 1) * -3 = -36\)
\(x_{max} = 12+1 = 13\)

Therefore, x can take \((13-7+1) = 7 \) different values.

Hence, the answer is E.

Correct answer : Choice C

Kinda guessing , but the logic being the range of numbers an-1 should be 5 since ax is between ax-1 -7 and ax-1 -2. Hence the number should be 5

A. 1
B. 4
C. 5 - Correct answer
D. 6
E. 7

a1 = -17

Because of the given equation, we can conclude that a2 will be one of -20, -21, -22, or -23.

Similarly, a3 will be one of -23, -24, -25, -26, -27, -28, -29.

For a4, it'll be one of -26, -27, -28, -29, -30, -31, -32, -33, -34, -35.

For a5, it'll be one of -29, -30, -31, -32, -33, -34, -35,..., -41.

For each additional term, the total term is increasing by 3. And for each additional term, the common possibilities are the total number of possibilities of the previous term minus 3.

So, if you extrapolate that, the first term that has -53 as a possibility is a7, which has a range of 19 numbers.
Each subsequent term's range has 3 lesser numbers in common with a7. So, given ax=-53, x has 6 possible values.

an-1 - 7 <an< an-1 -2

so a2 = -17-7<a2<-17-2

-24<a2<-19
so possible values of a2 are -23,-22.-21,-20
let a2 = -23, -30<a3<-25 so possible values of a3 are -29,-28,-27,-26
let a2 = -22 -29<a3<-24 so possible values of a3 are -28,-27,-26,-25
let a2= -21 -28<a3<-23 so possible values of a3 are -27,-26,-25,-24
let a2 = -20 -27<a3<-22 so possible values of a3 are -26,-25,-24,-23

Also
let a3 = -23 -30<a4<-25 so possible value of a4 are -29,-28,-27,-26
let a3 = -24 -31<a4<-26 so possible value of a4 are -30,-29,-28,-27
let a4 = -25 -32<-a4<-27 so possible values of a4 are -31,-30,-29,-28

as can be seen from the above pattern maximum time each value can repeat is 5. so I'll go with option C.

Answer: E

Infinite sequence of integers is a1, a2, a3, .... ,an

a1 = -17

a(n-1) - 7 < an < a(n-1) - 2 for n > 1

=> a1 - 7 < a2 < a1 - 2
Similarly, a2 - 7 < a3 < a2 - 2
a3 - 7 < a4 < a3 - 2
and so on...

a1 = -17, we get -24 < a2 < -19
i.e. a2 can have 4 values -23, -22, -21, -20.

To get the maximum range for a3, for LHS consider a2 = -23 and for RHS consider a2 = -20
=> -23-7 < a3 < -20 - 2
=> -30 < a3 < -22
i.e. a3 can have integer values -29, -28, -27,...., -24, -23

Similarly, -36 < a4 < -25
i.e. a4 can have integer values -35, -34, -33,....,-27, -26

For every next an, the range will increase by -6 on left side and increase by -3 on right side.

=> -36 < a4 < -25
=> -42 < a5 < -28
=> -48 < a6< -31
=> -54 < a7 < -34
=> -60 < a8 < -37
=> -66 < a9 < -40
=> -72 < a10 < -43
=> -78 < a11 < -46
=> -84 < a12 < -49
=> -90 < a13 < -52
=> -96 < a14 < -55
=> -102 < a15 < -58

Now for ax = −53, we can see from above that -53 will be in range for a8, a9, ... , a13 and a14
so x can take values 8, 9, 10, 11, 12, 13 and 14.
Total 7 values.

Given (an-1) - 7< an < (an-1) - 2 and a1 = -17
So, -24 < a2 < -19
Thus, considering all values of a2 (integer) we have a2 = {-23, -22, -21, -20}
So for a3 values would be a3 = {-29, ..., -23}
Similarly a4 = {-35, ..., -26}
Similarly a4 = {-41, ..., -29}
Thus we see a pattern that the starting term will have an integer which is previous starting minus 6 and the last term will be previous last term minus 3.
If we continue in this way we see that for ax = -53
criterion is being fulfilled from the term a7={-53,...,-35} upto term a13={-89,...,-53}
So different values that x can take are, 7,8,9,10,11,12 and 13. So total 7.
Hence answer choice E

A1 = -17
and other terms are governed by inequality An-1 -7 < An < An-1 - 2

we find that if n is odd An has 1 value and
if n is even than An has 4 different values

therefore, if Ax = -53, x is the 9th term
hence it can take 1 value
Answer A

Please refer to the attachment for the solution.

a1=-17
-24<a2<-19
-30<a3<-22
-36<a4<-25

Following the above trend, LHS is increasing by 6 and RHS is increasing by 3.
So,
-54<a7<-34
...<a13<-52
There are 7 numbers from a7 to a13.
Hence, answer is E.

I think the answer is 1.
For any given number, only 1 value is possible.

e.g. a2 will be between -19 to -24
for a2= -20, a3 will be between -22 to -27
for a3= -23, a4 will be between -25 to -30

If you see the pattern, there are no overlapping numbers. Hence the value of x will be 1.

Given: The infinite sequence of integers \(a_1\), \(a_2\), …, \(a_n\), … is such that \(a_1 = -17\) and \(a_{n-1} - 7 < a_n < a_{n-1} - 2\) for \(n > 1\).

Asked: If \(a_x=-53\), then how many different values can x take?

\(a_1 = -17\)
\(a_1 - 7 = -24 < a_2 < a_1 - 2 = -19\)
a_2 can take values = -23,-22,-21 or -20

Every successsive term is 3,4,5 or 6 less than previous term.

For max x,
-17 - 3(x-1) = - 53
3(x-1) = 53 -17 = 36
x = 13

For min x,
-17 - 6(x-1) = - 53
6(x-1) = 53 -17 = 36
x = 7

Since x is an integer, x can take value 7,8,9,10,11,12, or 13 : total 7 values.

IMO E

[quote="Bunuel"]The infinite sequence of integers \(a_1\), \(a_2\), …, \(a_n\), … is such that \(a_1 = -17\) and \(a_{n-1} - 7 < a_n < a_{n-1} - 2\) for \(n > 1\). If \(a_x=-53\), then how many different values can x take?

A. 1
B. 4
C. 5
D. 6
E. 7

Answer is E. Solution image attached

The infinite sequence of integers a1a1, a2a2, …, anan, … is such that a1=−17a1=−17 and an−1−7<an<an−1−2an−1−7<an<an−1−2 for n>1n>1. If ax=−53ax=−53, then how many different values can x take?

A. 1
B. 4
C. 5
D. 6
E. 7

Given, Ax = -53

Using the stimulus , we get

A(x-1) -7 < A(x) < A(x-1) -2

THus, x can take 4 values ranging from

A(x-1) -7 to A(x-1) -2 (both excluded)

[color=#0000ff](B) is the CORRECT answer[/color]

Answer is E

Posted from my mobile device

a1=-17 => -24 <a2< -19 then possible cases of a2=-23, -22, -21 and -20

if a3=-23 then -30 <a3< -25
if a3=-22 then -29 <a3< -24
if a3=-21 then -28 <a3< -23
if a3=-20 then -27 <a3< -22
all in all -30<a3<-22
left sides increased by 6 and right side increased by 3 then

-24 <a2<-19
-30<a3<-22
-36 <a4<-25
-42 <a5<-28
-50 <a6<-31
-56 <a7<-34
-62 <a8<-37
-68 <a9<-40
-74 <a10<-43
-80 <a11<-46
-86 <a12<-49
-92 <a13<-52
-98 <a14<-55
We can notice that x can take 7 values which are :7, 8, 9, 10, 11, 12 and 13

Answer is E