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

\(a_1 = -17\)

\(a_2\):
\(a_1-7 < a_2 <a_1-2\)
\(-17-7 < a_2 <-17-2\)
\(-24 < a_2 <-19\)
\(a_2\): {-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.

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

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.

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<a4<-28 {-32,-31,-30,-29} / {-35....-26} INC

a5: (-39,-34)(-38,-33)(-37,-32)(-36,-31) => a5=-35 / {-41....-29} INC

-42<a6<-37 {-41,-40,-39,-38} / {-47....-32} INC

a7: (-48,-43)(-47,-42)(-46,-41)(-45,-40) => a7=-44 / {-53....-35} YES[1]

-51<a8<-46 {-50,-49,-48, -47} / {-59....-38} YES[2]

a9: (-57,-52)(-56,-51)(-55,-50)(-54,-49) => a9=-53 (YES 1) / {-65....-41} YES[3]

-60<a10<-55 / {-71....-44} YES[4]
/ {-77....-47} YES[5]
/ {-83....-50} YES[6]
/ {-89....-53} YES[7]
/ {-95....-56}

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