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

Question

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

ThatDudeKnows · Accepted Answer

a_1 = -17

a_2 :
 a_1-7 < a_2 <a_1-2 
 -17-7 < a_2 <-17-2 
 -24 < a_2 <-19 
 a_2 : {-23, -22, -21, -20}

a_3 :
 -23-7 < a_3 <-20-2 
 -30 < a_3 <-22 
 a_3 : {-29, -28, -27, -26, -25, -24, -23}

Bottom end of the range moves to the left by 6 and top end of the range moves to the left by three each iteration.

a_4 : {-35,...-26}

a_5 : {-41,...-29}

a_6 : {-47,...-32}

a_7 : {-53,...-35}

a_8 : {-56,...-38}

a_9 : {,...-41}

a_9 : {,...-44}

a_10 : {,...-47}

a_11 : {,...-50}

a_12 : {,...-53}

That's seven.

Answer choice E.

Archit3110 · Answer

for given expression
a1=-17
a2 = -24<a2<-19
similarly solving for a3,a4,a5 
at a6 we get range for -53
option A is correct

sampathgelam · Answer

Given A1=-17 then putting the value of N in Inequality Equation for maximum value of N : -24-25 -52-27 -59-29 -66

paragamrutkar · Answer

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.

manish1708 · Answer

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.

av1901 · Answer

a1=-17 then as per the inequality of the question stem -24 < a2 < -19 => a2 = {-23,-22,-21,-20} Because sequence consists of integers only Now, if you consider a3 will have 4 different inequality ranges based on the value of a2 which will be: -30 < a3 < -25, -29 < a3 < -24, -28 < a3 < -23, -27 < a3 < -22 Combining the entire inequality will yield: -30 < a3 < -22 => a3 = {-29,-28,-27,-26,-25,-24,-23} All inclusive Using this logic, we can see that the inequality range of any term of the sequence can be determined by taking adding -6 to the highest term and -3 to the lowest term And the inclusive integers of the term can be determined by adding 1 to the lowest term of the inequality and subtracting 1 from the highest term of the inequality So, based on this logic, we can determine: a4 = {-35 to -26} All inclusive a5 = {-41 to -29} All inclusive a6 = {-47 to -32} All inclusive a7 = {-53 to -35} All inclusive : x=7 can = -53 (1) a8 = {-59 to -38} All inclusive : x=8 can = -53 (2) a9 = {-65 to -41} All inclusive : x=9 can = -53 (3) a10 = {-71 to -44} All inclusive : x=10 can = -53 (4) a11 = {-77 to -47} All inclusive : x=11 can = -53 (5) a12 = {-83 to -50} All inclusive : x=12 can = -53 (6) a13 = {-89 to -53} All inclusive : x=13 can = -53 (7) Now since -53 is the last term of a13, we know that following terms will be less than -53 in their entire range so no use going on from here We have 7 possible terms where the term can = -53 Answer - E -33 a5=-35 / {-41....-29} INC -42 a7=-44 / {-53....-35} YES -51 a9=-53 (YES 1) / {-65....-41} YES -60

wickedvikram · Answer

No. With this logic it will have 12 values of X. This would have been correct if the range were inclusive of both limits.

You need to displace by 1 in either side to be in the range, i.e. diff of 6 in the left and 3 in the right.

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

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

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