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APram
Joined: 23 Jun 2024
Last visit: 08 Sep 2026
Posts: 704
Location: India
Schools: ISB '26 HEC INSEAD Jan '26
GMAT Focus 1: 605 Q86 V78 DI76 
GPA: 3.608
a3 = a2 * (a1)^3 = 2 (a1)^3
a4 = a3 * (a2)^3 = 16 (a1)^3
a5 = a4 * (a3)^3 = 128 (a1)^12
With the given options and equations above, we can observe that a3 < a5
Also a3 can be negative but a5 is always positive
and multiplication factor is 128 for a5
So substituting the low value for a3 trying with -2
a1 = -1
a5 = 128
Hence this satisfies
We do not find any combination apart from this and is thus correctanswer
a3 = -2
a5 = 128
a4 = a3 * (a2)^3 = 16 (a1)^3
a5 = a4 * (a3)^3 = 128 (a1)^12
With the given options and equations above, we can observe that a3 < a5
Also a3 can be negative but a5 is always positive
and multiplication factor is 128 for a5
So substituting the low value for a3 trying with -2
a1 = -1
a5 = 128
Hence this satisfies
We do not find any combination apart from this and is thus correctanswer
a3 = -2
a5 = 128
Bunuel
Posts: 136
okay so
a3 = 2 * a1^3
a4 = 2 * a1^3 * a2^3 = 16 * a1^3
a5 = a4 * a3^3 = 16*a1^3 * 8*a1^9 = 128*a^12
Now if a1 = 1 then ans a3 = 2, a5 = 128
if a1 = -1 then ans a3 = -2 , a5 = 128
a3 = 2 * a1^3
a4 = 2 * a1^3 * a2^3 = 16 * a1^3
a5 = a4 * a3^3 = 16*a1^3 * 8*a1^9 = 128*a^12
Now if a1 = 1 then ans a3 = 2, a5 = 128
if a1 = -1 then ans a3 = -2 , a5 = 128
Posts: 36
a5 = a4 * (a3)^3;
a4 = a3 * (a2)^3;
Plug-in the value a3 = 4;
a4 = 4 * 8 = 32
a5 = 32 * 64; (Eliminate)
Plug-in the value a3 = -2;
a4 = -2 * 8 = -16
a5 = -16 * -8 = 128;
Therefore,
a3 = -2; Option E
a5 = 128; Option A
a4 = a3 * (a2)^3;
Plug-in the value a3 = 4;
a4 = 4 * 8 = 32
a5 = 32 * 64; (Eliminate)
Plug-in the value a3 = -2;
a4 = -2 * 8 = -16
a5 = -16 * -8 = 128;
Therefore,
a3 = -2; Option E
a5 = 128; Option A
Dav2000
Joined: 21 Sep 2023
Last visit: 03 Jan 2026
Posts: 75
Location: India
Schools: Full Time MBA '26 (A)
GPA: 8.6
Schools: Full Time MBA '26 (A)
Posts: 75
Bunuel
given a2 = 2
and a(n) = a(n-1)*(a(n-2))^3
Therefore, a3 = a2 * ((a1)^3) = 2* (a1)^3
and a4 = a3 * (a2) ^ 3 = 16 * (a1)^3
and a5 = a4 * (a3) ^3 = 32 * (a1)^6, Since there is only one option that is a multiple of 32 Therefore a5= 128. Option A or 128
And from a5 = 128 = 32 * (a1)^6 we can get the value of a1 = 2^(1/3)
then the value of a3 = 2 *(2 ^(1/3)) ^3 = 4. Hence Option D or Option 4
Bunuel
Solved using hit and trial method of options
asuming a3=-8
-8 = 2 * a1 ^3 , so a1 = -1
and therefore a5 = a4 * a3^3 = 128
putting all other options as a3 produces non integer
Posts: 198
from above equation
a3=a2*a1^3= 2a1^3
a4=a3*a2^3=8a3
a5=a4*a3^3=8a3^4
lets take a1=2
a3=4
a5=8*16*16
not possible
a1=1
a3=2
a5=128
Not possible
a1=-1
a3=-2
a5=128
Correct answer is a3=-2 & a5=128
a3=a2*a1^3= 2a1^3
a4=a3*a2^3=8a3
a5=a4*a3^3=8a3^4
lets take a1=2
a3=4
a5=8*16*16
not possible
a1=1
a3=2
a5=128
Not possible
a1=-1
a3=-2
a5=128
Correct answer is a3=-2 & a5=128
Rakshit25
Joined: 15 Jun 2020
Last visit: 23 Mar 2026
Posts: 120
Location: India
Schools: INSEAD Aug '26 (A) LBS '26 (D) IESE '26 (A)
GPA: 8.4
Schools: INSEAD Aug '26 (A) LBS '26 (D) IESE '26 (A)
Posts: 120
Made two equations basically:
a4=a3*2^3=8(a3)
a5=8(a3)*a3
Only a3=8 and subsequently a5=128 satisfy that equation. Hence these two should be the options
a4=a3*2^3=8(a3)
a5=8(a3)*a3
Only a3=8 and subsequently a5=128 satisfy that equation. Hence these two should be the options
Bunuel
Posts: 19
a3=(a2) X (a1)^3
a3= 2(a1)^3
Similarly,
a5=(a4) X (a3)^3
a5=((a3) X (a2)^3) X (2(a1)^3)^3
a5= (2(a1)^3 X 8) X 8(a1)^9
a5= 128(a1)^12
a5=(2^7) X (a1)^12
a5=(2^3) X [2(a1)^3]^4
now substituting a3 in this equation,
a5=8(a3).
So among the options,
only values for a3 and a5 that satisfy the above equation are 16 & 128 respectively.
So, a3=16 and a5=128
a3= 2(a1)^3
Similarly,
a5=(a4) X (a3)^3
a5=((a3) X (a2)^3) X (2(a1)^3)^3
a5= (2(a1)^3 X 8) X 8(a1)^9
a5= 128(a1)^12
a5=(2^7) X (a1)^12
a5=(2^3) X [2(a1)^3]^4
now substituting a3 in this equation,
a5=8(a3).
So among the options,
only values for a3 and a5 that satisfy the above equation are 16 & 128 respectively.
So, a3=16 and a5=128
MaxFabianKirchner
Joined: 02 Jun 2025
Last visit: 12 Jul 2025
Posts: 65
Status:26' Applicant
Location: Denmark
Concentration: Finance, International Business
Schools: ESSEC MiF '26 ESADE HEC Imperial LBS Oxford MFE '26 Sloan (MIT) NYU Stern NUS Bocconi IE'26
GPA: 4.0
WE:Other (Student)
Schools: ESSEC MiF '26 ESADE HEC Imperial LBS Oxford MFE '26 Sloan (MIT) NYU Stern NUS Bocconi IE'26
Posts: 65
We are given the recurrence relation a(n) = a_(n-1) * (a_(n-2))^3 for n > 2, with a_2 = 2. The term a(1) must be a non-zero integer.
First, let's express the terms we need, a_3 and a_5, using the given information.
For n = 3, the formula gives:
a_3 = a_2 * a_1^3
Since a(2) = 2, this becomes:
a_3 = 2 * (a_1)^3
Next, let's derive an expression for a_5. We first need a_4.
For n = 4: a_4 = a_3 * (a_2)^3 = a_3 * (2)^3 = 8 * a_3.
For n = 5: a_5 = a_4 * a_3^3.
Substituting our expression for a_4, we get:
a_5 = (8 * a_3) * (a_3)^3 = 8 * (a_3)^4.
So, we have two key equations:
1. a_3 = 2 * (a_1)^3
2. a_5 = 8 * (a_3)^4
The first equation tells us that a_1 is the cube root of a_3/2. Since a_1 must be a non-zero integer, we can test the options provided for a_3 to see which one yields an integer value for a_1.
Let's test the option a_3 = -2.
If a_3 = -2, then (a_1)^3 = -2 / 2 = -1. The cube root of -1 is -1, so a_1 = -1. This is a valid non-zero integer.
Now, we use this value of a_3 to find the corresponding value of a_5 using our second equation.
If a_3 = -2, then a_5 = 8 * (-2)^4 = 8 * 16 = 128.
The pair of values a_3 = -2 and a_5 = 128 are both present in the options list and are consistent with the sequence's rules, originating from a_1 = -1. Checking other options for a_3 will not produce a value for a_5 that is also in the list.
Therefore, the correct value for a_3 is -2 and for a_5 is 128.
First, let's express the terms we need, a_3 and a_5, using the given information.
For n = 3, the formula gives:
a_3 = a_2 * a_1^3
Since a(2) = 2, this becomes:
a_3 = 2 * (a_1)^3
Next, let's derive an expression for a_5. We first need a_4.
For n = 4: a_4 = a_3 * (a_2)^3 = a_3 * (2)^3 = 8 * a_3.
For n = 5: a_5 = a_4 * a_3^3.
Substituting our expression for a_4, we get:
a_5 = (8 * a_3) * (a_3)^3 = 8 * (a_3)^4.
So, we have two key equations:
1. a_3 = 2 * (a_1)^3
2. a_5 = 8 * (a_3)^4
The first equation tells us that a_1 is the cube root of a_3/2. Since a_1 must be a non-zero integer, we can test the options provided for a_3 to see which one yields an integer value for a_1.
Let's test the option a_3 = -2.
If a_3 = -2, then (a_1)^3 = -2 / 2 = -1. The cube root of -1 is -1, so a_1 = -1. This is a valid non-zero integer.
Now, we use this value of a_3 to find the corresponding value of a_5 using our second equation.
If a_3 = -2, then a_5 = 8 * (-2)^4 = 8 * 16 = 128.
The pair of values a_3 = -2 and a_5 = 128 are both present in the options list and are consistent with the sequence's rules, originating from a_1 = -1. Checking other options for a_3 will not produce a value for a_5 that is also in the list.
Therefore, the correct value for a_3 is -2 and for a_5 is 128.
Bunuel
We know that \(a_2 = 2\). Basis the expression given, we can write:
\( a_3 = a_2 * a_1^3 = 2 * a_1^3\)
Similarly, we can write:
\(a_4 = a_3 * a_2^3 = 2^4 * a_1^3\)
\(a_5 = a_4 * a_3^3 = 2^7 * a_1^1^2\)
Looking at just the expression we got for \(a_3\) and \(a_5\),
We can say that if \(a_3 = m\), then \(a_5 = 8m^4\)
Now we just need to find a value for \(m\), that satisfies the values for both \(a_3\) and \(a_5\),
At, \(m = -2\), we get \(8m^4 = 128\)
Thus, these are the correct corresponding values.
\(a_3 = -2\)
\(a_5 = 128\)
\( a_3 = a_2 * a_1^3 = 2 * a_1^3\)
Similarly, we can write:
\(a_4 = a_3 * a_2^3 = 2^4 * a_1^3\)
\(a_5 = a_4 * a_3^3 = 2^7 * a_1^1^2\)
Looking at just the expression we got for \(a_3\) and \(a_5\),
We can say that if \(a_3 = m\), then \(a_5 = 8m^4\)
Now we just need to find a value for \(m\), that satisfies the values for both \(a_3\) and \(a_5\),
At, \(m = -2\), we get \(8m^4 = 128\)
Thus, these are the correct corresponding values.
\(a_3 = -2\)
\(a_5 = 128\)
We are told that each number in the sequence depends on the two numbers before it, and that the second number is 2. To find which values of the third and fifth numbers fit this rule, we can start by guessing a value for the first number. If we assume the first number is -1, then the third number comes out to be -2 using the given rule. Continuing the pattern using these values, the fifth number works out to be 128. This matches the choices in the table, so the correct pair that follows the rule is the third number as -2 and the fifth number as 128.
Bunuel
Mohak01
Joined: 05 Sep 2020
Last visit: 01 Apr 2026
Posts: 104
Location: India
Schools: ISB '26 NUS '26 (A) IIMA '26 (WL) EPGP '26 (D)
GMAT Focus 1: 695 Q83 V87 DI83 
GPA: 8
Given a2 = 2, using given equation, following values can be calculated in terms of a1
a3 = 2*(a1^3)
a4= 16*(a1^3)
a5 = 128*(a1^12)
Substituting some random values of a1 such as 1,-1 and 1/2,
Only a1= -1, a3=-2 and a5 = 128 satisfies the above equations.
a3 = 2*(a1^3)
a4= 16*(a1^3)
a5 = 128*(a1^12)
Substituting some random values of a1 such as 1,-1 and 1/2,
Only a1= -1, a3=-2 and a5 = 128 satisfies the above equations.
Posts: 88
We will determine possible values for a3:
n=3,
a3=a2*(a1)^3=2*(a1)^3
We will look for form 2k^3 from answer choices:
valid a3 candidates are 128, 16, and −2.
Now, for a5:
a4=a3*(a2)^3=> a3*8
a5=a4*(a3)^3
From answer choices:
If a3=−2,
a4=−2*8=−16
a5=−16*(−2)^3
=−16*(−8)=> 128
Therefore, a3=-2 and a5=128.
n=3,
a3=a2*(a1)^3=2*(a1)^3
We will look for form 2k^3 from answer choices:
valid a3 candidates are 128, 16, and −2.
Now, for a5:
a4=a3*(a2)^3=> a3*8
a5=a4*(a3)^3
From answer choices:
If a3=−2,
a4=−2*8=−16
a5=−16*(−2)^3
=−16*(−8)=> 128
Therefore, a3=-2 and a5=128.
Posts: 731
Bunuel
We know that a5 = a4 (a3)^3 as per the given equation.
We also know that a4 = a3(a2)^3 => 8a3. So if we substitute this a5 = 8a3 (a3)^3 => a5 = 8(a3)^4.
So we need to find the matching pair \(a5 = 8(a3)^4\). If a3 = -2, a5 = 128.
Hence IMO a3 = -2 , a5 = 128.
Praveena_10
Joined: 01 Jan 2024
Last visit: 24 Aug 2026
Posts: 46
Location: India
Schools: INSEAD Jan '27 ISB '27 HEC Sept '26
GPA: 7.3
WE:Engineering (Energy)
Schools: INSEAD Jan '27 ISB '27 HEC Sept '26
Posts: 46
let a1=x
a3=a2(a1)3=2.x3
a5=a4(a3)3=a3(a2)3(a3)3
=(a3)4*8=128*x12
let us assume a1=1,
a3=2, a5= 128--No
let us assume a1 as -1;
a3=-2, a5=128.
Hence this will be answer.
a3=a2(a1)3=2.x3
a5=a4(a3)3=a3(a2)3(a3)3
=(a3)4*8=128*x12
let us assume a1=1,
a3=2, a5= 128--No
let us assume a1 as -1;
a3=-2, a5=128.
Hence this will be answer.
Bunuel
We have to ensure the unknown value as x form and equation then use the options to solve it.
Given:
- a2 = 2
- Formula: an = a(n−1) * (a(n−2))^3 for n > 2
- Let a1 = X
- a3 = 2 * X^3
- a4 = a3 * 2^3 = (2 * X^3) * 8 = 16 * X^3
- a5 = a4 * (a3)^3 = (16 * X^3) * (2 * X^3)^3
= 16 * X^3 * 8 * X^9 = 128 * X^12
Now try values:
If X = -1:
- a3 = 2 * (-1)^3 = -2
- a5 = 128 * (-1)^12 = 128
Hence :
a3 = -2, a5 = 128
Posts: 32
a2=2 => a3=2 x (a1)^3 => a3 should be a number that (a3/2) is a cubic of an integer (because a1 is an integer) => Only 16 and -2 is reasonable for being a3 (16/2=8=2^3 and -2/2=-1=(-1)^3)
If a3=16 => a4=16x2^3=16x8=128 => a5= 128x16^3 => So big for any other options to choose.
If a3=-2 => a4=(-2)x2^3=-16 => a5=(-16)x(-2)^3=1288 => a3=-2 and a5=128
If a3=16 => a4=16x2^3=16x8=128 => a5= 128x16^3 => So big for any other options to choose.
If a3=-2 => a4=(-2)x2^3=-16 => a5=(-16)x(-2)^3=1288 => a3=-2 and a5=128
Bunuel
Bunuel
The sequence is mentioned in the question stem for n>2
\(a_n = (a_{n-1})(a_{n-2})^3\) for \(n > 2\), and \(a_2 = 2\).
a2= 2
a3 = a2 * (a1)^3 = 2*(a1)^3
a4 = a3 * (a2)^3 = 2*(a1)^3 * (2)^3 = 16 (a1)^3
a5 = a4 * (a3)^3 = 16* (a1)^3 * (2*(a1)^3)^3 = 128 (a1)^12
a1 cannot be 1. Then the values of a3 and a5 becomes 2, 128.
if we take a1 = -1 ( non zero integers ) , then a3 = -2 and a5 = 128
Thus, the values of a3 = -2 , a5 = 128.
Posts: 67
Given :
an=(an−1)(an−2)^3 for n>2 and an = 2
Let’s assume a1=1 or -1
Assume a1=1:
Assume a1=-1:
ans: a3 = -2, a5 = 128
an=(an−1)(an−2)^3 for n>2 and an = 2
Let’s assume a1=1 or -1
Assume a1=1:
- a2=2 (given)
- a3=2*(1)^3=2
- a4=2*(2)^3=2*8=16
- a5=16*(2)^3=16*8=128
Assume a1=-1:
- a2=2 (given)
- a3=2*(-1)^3=-2
- a4=-2*(2)^3=-2*8=-16
- a5=-16*(-2)^3=-16 (-8) =128
ans: a3 = -2, a5 = 128
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