Last visit was: 30 Apr 2026, 12:43

It is currently 30 Apr 2026, 12:43

Toolkit

Practice Tests

Data Insights Questions

Two-part Analysis (TPA)

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Go to My Error Log Learn more

Request Expert Reply

Confirm Cancel

Back to Forum

Create Topic

GMAT Club Olympics 2025 (Day 4): The infinite sequence of nonzero

1 2 3 4 5 6

ODST117

Joined: 15 Aug 2024

Last visit: 27 Jan 2026

Posts: 173

Own Kudos:

[1]

Given Kudos: 149

Posts: 173

Kudos: 88

[1]

04 Jul 2025, 04:18

Kudos

Bookmarks

$a_1, a_2, a_3$..... are all nonzero integers and are defined $a_n = (a_{n-1})(a_{n-2})^3$ where $n>2$ and $a_2=2$

$a_3=a_2(a_1)^3$
$a_3=2(a_1)^3$

$a_4=a_3(a_2)^3$
$a_4 = a_2(a_1)^3(2)^3$
$a_4 =16 (a_1)^3$

We have $(a_1)^3$ in both $a_3$ and $a_4$. Find $(a_1)^3$ in terms of $a_3$ and $a_4$ and set them equal to eachother

$(a_1)^3 = \frac{a_3}{2}$

$(a_1)^3 = \frac{a_4}{16}$

$\frac{a_3}{2} = \frac{a_4}{16}$
$a_4 = 8a_3$

$a_5 = a_4(a_3)^3$
$a_5 = 8a_3(a_3)^3$
$a_5 = 8(a_3)^4$

When $a_3 = -2$, $a_5 = 128$

quynhanhtran

Joined: 25 Jan 2018

Last visit: 23 Feb 2026

Posts: 38

Own Kudos:

[1]

Given Kudos: 62

Posts: 38

Kudos: 28

[1]

04 Jul 2025, 04:36

Kudos

Bookmarks

Let a1 = x, a2=2
Then:
a3 = a2 * (a1)^3 = 2x^3
a4 = a3 * (a2)^3 = (2x)^3 * 8 = 16x^3
a5 = a4 * (a3)^3 = (16*x^3)(2*x^3)^3 = 16*x^3 * 8*x^9 = 128 * x^12
=> a5 = 128*x^12

x^12 is larger or equal to 0 => a^5 is larger or equal 128 => a^5 = 128(among options in column)
=> a1 may be -1 or 1
=> a3 maybe -2 or 2.
There's no 2 as an option, so a3 = -2

Answer: a3 = -2, a5 = 128

SaanjK26

Joined: 08 Oct 2022

Last visit: 30 Apr 2026

Posts: 77

Own Kudos:

[1]

Given Kudos: 76

Location: India

Posts: 77

Kudos: 63

[1]

04 Jul 2025, 04:40

Kudos

Bookmarks

recursive rule: a_n =(a_n-1)(a_n-2)^3 and a_2 = 2

Using the rule, we express a_(3) = 2.a_1 ^ 3 and a_(5) = 8a_(3) ^ 4

Testing a_(1) = - 1
a_(3) = - 2.

Plugging into the formula, a_(5) = 8 * (- 2) ^ 4 = 8×16 128.

Thus,a_(3) = - 2 and a_(5) = 128.

crimson_king

Joined: 21 Dec 2023

Last visit: 30 Apr 2026

Posts: 152

Own Kudos:

156

[1]

Given Kudos: 113

GRE 1: Q170 V170 Verified score

Posts: 152

Kudos: 156

[1]

04 Jul 2025, 04:41

Kudos

Bookmarks

If a2=2 & an = (an-1)*(an-2)^3

If we assume a1 to be 1 then a3 would be 2 but this is not among the options given.

Now if we assume a1 to be -1 then a3 = 2* (-1)^3 = -2. This is one of the options hence if we assume the value of a3 = -2.

Then if we calculate the value of a4, we get a4=-2*(2)^3 = -16

Now if we calculate the value of a5, we get a5=-16*(-2)^3 = 128

Hence the values of a3 & a5 given are :
a3 = -2
a5= 128

UfuomaOh

McCombs School Moderator

Joined: 14 Sep 2023

Last visit: 24 Apr 2026

Posts: 83

Own Kudos:

[1]

Given Kudos: 14

Posts: 83

Kudos: 50

[1]

04 Jul 2025, 04:43

Kudos

Bookmarks

from the expression

a3=(a2)(a1)^3
a4=(a3)(a2)^3
a5=(a4)(a3)^3
a5=(a3)(a2)^3(a3)^3
a5=(a3)^4. (a2)^3
a2=2
a5=8.(a3)^4
(a3)^4 is positive given the power of the indices

a3=(a2)(a1)^3
a3= 2.(a1)^3
So a3 can be either positive or negative given the power of the indices.
a3 is a multiple of 2 which makes a5 a multiple of 2

Since we don't know a1 and we have options we could plug and play

If we assume a5 is 128. Then

128= 8*(a3)^4
(a3)^4= 64

(a3)^(4*1/4)= 64^1/4
a3= + or - 2

shaliny

Joined: 30 Oct 2023

Last visit: 30 Apr 2026

Posts: 121

Own Kudos:

[1]

Given Kudos: 1,014

Products:

Posts: 121

Kudos: 26

[1]

04 Jul 2025, 04:46

Kudos

Bookmarks

a4= a3 a2^3 a4= a3 2^3(since a2 = 2 ; given) a4 = 8a3 a5= a4 a3^3 a5 = 8a3 a3^3 = 8*a3^4 or, a5/8 = a3^4
so by this we can interpret that we need a number for a5 which is divisible by 8 and for a3 a number could be square root 4 of a number after dividing by 8 so, in the given table the only numbers that satisfy this is- 128 for a5 and -2 for a3.

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

The infinite sequence of nonzero integers $a_1$, $a_2$, $a_3$, ... is such that $a_n = (a_{n-1})(a_{n-2})^3$ for $n > 2$, and $a_2 = 2$.

Select for $a_3$ and $a_5$ the values of $a_3$ and $a_5$, respectively, that would be jointly consistent with the given information. Make only two selections, one in each column.

SaKVSF16

Joined: 31 May 2024

Last visit: 29 Apr 2026

Posts: 86

Own Kudos:

[1]

Given Kudos: 41

GMAT Focus 1: 695 Q87 V82 DI85 Verified score

GPA: 3.6

Products:

GMAT Focus 1: 695 Q87 V82 DI85 Verified score

Posts: 86

Kudos: 80

[1]

04 Jul 2025, 04:57

Kudos

Bookmarks

According to the formula given

$\\
a_3 = a_2 * (a_1)^3 = 2(a_1)^3\\
a_4 = a_3 * (a_2)^3 = 8(a_3)\\
a_5 = a_4 * (a_3)^3 = 8(a_3)* (a_3)^3 = 8 * (a_3)^4 = 8 *(2(a_1)^3 )^4 = 128 * (a_1)^{12}\\
$

so $a_5 = 128 (a_1)^{12} $ and $ a_3= 2(a_1)^3 $

we need to find a value for $a_1$ which would give an $a_3 $ and $a_5$ that matches the options given to us.
We see 128 and -2 in the options which can be yielded if $a_1$ = -1

$a_1= -1$
then $a_5 = 128 (-1)^{12} = 128$
and $a_3= 2(-1)^3 = -2 $

select -2 for $a_3 $ and 128 for $a_5$

Rahilgaur

Joined: 24 Jun 2024

Last visit: 26 Jan 2026

Posts: 162

Own Kudos:

125

[1]

Given Kudos: 47

GMAT Focus 1: 575 Q81 V82 DI72 Verified score

Products:

GMAT Focus 1: 575 Q81 V82 DI72 Verified score

Posts: 162

Kudos: 125

[1]

04 Jul 2025, 05:01

Kudos

Bookmarks

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

Show Spoiler

Attachment:

GMAT-Club-Forum-1k8zqyjm.jpeg [ 202.67 KiB | Viewed 278 times ]

jnxci

Joined: 29 Nov 2018

Last visit: 28 Apr 2026

Posts: 39

Own Kudos:

[1]

Given Kudos: 338

Posts: 39

Kudos: 18

[1]

04 Jul 2025, 05:46

Kudos

Bookmarks

a(4) = a(3)*a(2)^3
a(5) = a(4)*a(3)^3 = a(3)^4*a(2)^3
--> a(5) = 2^3*a(3)^4
--> Only a(3) = -2 & a(5) =8*16 = 128 works

GMAT1034

Joined: 21 Feb 2023

Last visit: 29 Apr 2026

Posts: 265

Own Kudos:

[1]

Given Kudos: 154

Products:

Posts: 265

Kudos: 65

[1]

04 Jul 2025, 05:50

Kudos

Bookmarks

Just keep on putting the values, since n>2, put n=3, 4, 5. Try to convert all of these in one form as in a3 in terms of a2, a1, a4 in terms of a2, a1, and a5 in terms of a2, a1. Why that because you have the value of a2.
Doing that you will get a3= 2( a1)^3
and a5= 126. a1^6.
We dont know a1 now, but going through the answer choices, it can be realised that a1 has to be -1. And then , a3=-2, a5=128

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

LastHero

Joined: 15 Dec 2024

Last visit: 12 Dec 2025

Posts: 134

Own Kudos:

147

[1]

Given Kudos: 1

Posts: 134

Kudos: 147

[1]

04 Jul 2025, 05:55

Kudos

Bookmarks

a3 = 2 * a1^3
a4 = a3 * 2^3 = 2 * a1^3 * 2^3 = 2^4 * a1^3
a5 = a4 * a3^3 = 2^4 * a1^3 * (2 * a1^3)^3 = 2^4 * a1^3 * 2^3 * a1^9 = 2^7 * a1^12

In the options, only 128 "has" 2^7 inside, a5=128
And a1 can be 1 or -1.
If a1=1, a3=2, not in the options
If a1=-1, a3=-2, in the options

a3=-2 and a5=128

SumnerSCB

Joined: 27 Apr 2025

Last visit: 08 Sep 2025

Posts: 36

Own Kudos:

[1]

Posts: 36

Kudos: 22

[1]

04 Jul 2025, 06:17

Kudos

Bookmarks

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

So for a3 we get it by (a2)(a1)^3 let a1 be x because we don't know it. 2x^3 is a3 we can then test out different values to work up the chain to a5. We see based on the formula a5 must be much larger than a3 so for values of a3 I would start at the low end. -8. then for a4 (-8)(2)^3 or -8(8) which would be -64 so for a5 we get (-64)(-8)^3 to large. So a3 cannot be -8. if a3 = -2 then a4=(-2)(2)^3 or -16 than a5 would be (-16)(-2)^3 or -16*8 or 128. which is an option, so done.

missionmba2025

Joined: 07 May 2023

Last visit: 07 Sep 2025

Posts: 341

Own Kudos:

431

[1]

Given Kudos: 52

Location: India

Posts: 341

Kudos: 431

[1]

04 Jul 2025, 06:23

Kudos

Bookmarks

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

As ^1 cannot be zero, it can be -1

Final Answers

a^3 = -2
a^6 = 128

Show Spoiler

Attachment:

GMAT-Club-Forum-i091bla4.jpeg [ 71.44 KiB | Viewed 258 times ]

RedYellow

Joined: 28 Jun 2025

Last visit: 09 Nov 2025

Posts: 80

Own Kudos:

[1]

Posts: 80

Kudos: 74

[1]

04 Jul 2025, 06:39

Kudos

Bookmarks

Testing the values, the correct ones are a3=-2 and a5=128
a2=2, so a4 will have a 2^3 and a5 at least another 2^3. There is at least a 2^6 in the answer of a5. a5=128=2^7 is a good canditate.
The other 2 of a5 comes because a3=-2.

Mardee

Joined: 22 Nov 2022

Last visit: 02 Feb 2026

Posts: 225

Own Kudos:

191

[1]

Given Kudos: 20

Posts: 225

Kudos: 191

[1]

04 Jul 2025, 06:50

Kudos

Bookmarks

All the sequences are non zero integers
a2 = 2
an = (an-1)(an-2)^3

Let us consider a1 = x
From the given recurrence,
a3 = a2*a1^3 = 2x^3
a4 = a3*(a2)^3 = (2x^3)(2^3) = 16x^3
a5 = a4*(a3)^3 = (16x^3)*(2x^3)^3 = 128x^12

Lets try from the lowest number for a3

a3 = -8
=> -8 = 2x^3
=> -4 = x^3 (Not possible since x not an integer here)

a3 = -2
=> -2 = 2x^3
=> -1 = x^3
=> x = -1

So, a5 = 128(-1)^12 = 128 which matches with the given options

a3 = -2
a5 = 128

Gg153

Joined: 25 Oct 2024

Last visit: 02 Feb 2026

Posts: 99

Own Kudos:

[1]

Given Kudos: 102

Location: India

Products:

Posts: 99

Kudos: 52

[1]

04 Jul 2025, 06:53

Kudos

Bookmarks

$a_3 = 2 * (a_1)^3 $
$a_4 = 2 * (a_1)^3 * 8 $
$a_5 = 16 * (a_1)^3 * 8* (a_1)^9 $
so we find answers where $a_3 = 2 * (a_1)^3 $ and $a_5 = 128 * (a_1)^1^2 $
This only works in the options when $a_1 = -1$ so $a_3 = -2$ and $a_5 = 128$

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

Lemniscate

Joined: 28 Jun 2025

Last visit: 09 Nov 2025

Posts: 80

Own Kudos:

[1]

Posts: 80

Kudos: 72

[1]

04 Jul 2025, 07:11

Kudos

Bookmarks

The number with the smallest absolute value is -2, which is a good aspirant to be a3 (raising a number to the third power makes it bigger quickly).
-2=2*(a1)^3
a1=-1

a4 = -2 * 2^3 = -16
a5 = -16 * (-2)^3 = 128

Solutions: a3 is -2 and a5 is 128

samriddhi1234

Joined: 19 Aug 2022

Last visit: 09 Jan 2026

Posts: 45

Own Kudos:

[1]

Given Kudos: 303

Location: India

Schools: Rotterdam'26

GMAT Focus 1: 625 Q80 V85 DI78 Verified score

GPA: 9.17

Schools: Rotterdam'26

GMAT Focus 1: 625 Q80 V85 DI78 Verified score

Posts: 45

Kudos: 25

[1]

04 Jul 2025, 07:12

Kudos

Bookmarks

Using the options, checking one by one, the answer is a3 = -2 and a5 = 128.

$ a3 = a2*(a1)^3 = 2*(a1)^(3) $

$ a5 = a4*(a3)^3 = (a3*(a2)^3)*(a3)^3 = 128*(a1)^(12) $

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

Aarushi100

Joined: 17 Jul 2024

Last visit: 26 Apr 2026

Posts: 61

Own Kudos:

[1]

Given Kudos: 96

Posts: 61

Kudos: 49

[1]

04 Jul 2025, 07:20

Kudos

Bookmarks

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

Given to us:
a1,a2,a3..... infinite sequence of nonzero integers.
[ltr]an=(an−1)(an−2)^3 for n>2
a2=2

Now, a3=?
a5=?

For a3 we need a1.
Let a1=x
a3= a2*a1^3
= 2x^3

a4= a3*a2^3
= 2x^3*(2)^3
=16x^3

a5= a4*a3^3
= 16x^3*(2x^3)^3
= 128x^12

Now, let's look at the options for probable values of x.
The value of a3 must be small as the maximum possible value of a5 can be 128.
And according to our calculation, a5= 128x^12
=> x=-1 or 1
But, if x=1, then a3=2 isn't an option.
Thus, x=-1. => a3=-2

And
128(-1)^12= a5= 128

a3=-2
a5=128

[/ltr]

adityaprateek15

Joined: 26 May 2023

Last visit: 27 Apr 2026

Posts: 346

Own Kudos:

170

[1]

Given Kudos: 323

Location: India

Schools: Ross '26 Tepper '26 McCombs '26

GPA: 2.7

Schools: Ross '26 Tepper '26 McCombs '26

Posts: 346

Kudos: 170

[1]

04 Jul 2025, 07:37

Kudos

Bookmarks

[ltr]Given,
an=(an−1)(an−2)^3 for n>2
and a2 = 2

So, a3 = a2*(a1)^3
a3= 2*(a1)^3

a4= a3*(a2)^3
2(a1)^3*2^3
2*(a1)^3*8
16(a1)^3

a5= a4*(a3)^3
[16(a1)^3]*[ 2*(a1)^3]^3
16(a1)^3*8*(a1)^9
128*(a1)^12

So, a3 = 2*(a1)^3
a5 = 128*(a1)^12

Looking at the option choices, we don't have any possible options for a5 = 128*(a1)^12 -1=<a1>=1
Suppose a1=2, a5 = 128*2^12. Not there in the options.

if a1 = 1, a3 = 2 (not there in options)
if a1 = -1, a3 = -2 (Yes) and a5 = 128(-1)^12 = 128

Hence, a3 = -2 and a5 = 128[/ltr]

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more