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04 Jul 2025, 07:49
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It's key to not confuse yourself and keep your calculations neat for a question like this.
If A3= -2
and
A2=2 the
A4= (-2)(2)^3=-2*8=-16
A5=-16*(-2)^3=-16*-8=128
If A3= -2
and
A2=2 the
A4= (-2)(2)^3=-2*8=-16
A5=-16*(-2)^3=-16*-8=128
Bunuel
04 Jul 2025, 07:49
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First, let's find a way to express a5 using a3 :
a3 =a2 * (a1)^3 = 2*(a1)^3
a4= a3*(a2)^3 =a3 (2^3) =8*a3
a5 = a4*(a3)^3 = (8*a3)(a3)^3 = 8(a3)^4
Now, we need to find values for a3 and a5 from the options that fit two conditions:
a1 must be a nonzero whole number (because the sequence is made of nonzero integers).
This means a3 must be twice the cube of a whole number (e.g., 2×1^3 = 2, 2×(−1)^3 = −2, etc.).
The calculated a5 must match one of the given a5 options.
Let's check the options for a3:
If a3 =1 : (a1)^3 =1/2. No whole number for a1
If a3 = 2: a1 =1. This works!
If a3 =2, then a5 = 8*(2)^4 =8×16=128. This a5 is in the options!
So, both a3 =2 and a3 =−2 are possible, and they both lead to a5 =128.
Since 128 is the only a5 value that works, and 2 is a valid choice for a3 , I'll go with that.
Answer is:
a3 = -2 & a5 =128
a3 =a2 * (a1)^3 = 2*(a1)^3
a4= a3*(a2)^3 =a3 (2^3) =8*a3
a5 = a4*(a3)^3 = (8*a3)(a3)^3 = 8(a3)^4
Now, we need to find values for a3 and a5 from the options that fit two conditions:
a1 must be a nonzero whole number (because the sequence is made of nonzero integers).
This means a3 must be twice the cube of a whole number (e.g., 2×1^3 = 2, 2×(−1)^3 = −2, etc.).
The calculated a5 must match one of the given a5 options.
Let's check the options for a3:
If a3 =1 : (a1)^3 =1/2. No whole number for a1
If a3 = 2: a1 =1. This works!
If a3 =2, then a5 = 8*(2)^4 =8×16=128. This a5 is in the options!
So, both a3 =2 and a3 =−2 are possible, and they both lead to a5 =128.
Since 128 is the only a5 value that works, and 2 is a valid choice for a3 , I'll go with that.
Answer is:
a3 = -2 & a5 =128
04 Jul 2025, 07:53
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Assume a1 = 1 given a2= 2 then a3 =2x1= 2 but we have no 2
Assume a1 is = -1 given a2= 2 then a3 = 2x-1= -2, a4 = 2x-8= -16 and a5= -16x-8 = 128
a3= -2 and a5= 128
Assume a1 is = -1 given a2= 2 then a3 = 2x-1= -2, a4 = 2x-8= -16 and a5= -16x-8 = 128
a3= -2 and a5= 128
Bunuel
