What is a formula for the nth term of the given sequence?

50, comma, minus, 10, comma, 2, point, point, .
50,−10,2...

1 answer

To find a formula for the nth term of the sequence \(50, -10, 2\), we first observe the pattern.

Let's denote the sequence as \(a_n\), where:

  • \(a_1 = 50\)
  • \(a_2 = -10\)
  • \(a_3 = 2\)

Next, we analyze the differences between consecutive terms:

  • \(a_2 - a_1 = -10 - 50 = -60\)
  • \(a_3 - a_2 = 2 - (-10) = 2 + 10 = 12\)

The differences are:

  • From \(50\) to \(-10\) is \(-60\)
  • From \(-10\) to \(2\) is \(12\)

Now, let's look for a potential pattern or function. It appears that the numbers are alternating in behavior and magnitude. A quadratic function could potentially represent this sequence.

Let's assume the general term is of the form: \[ a_n = An^2 + Bn + C \]

Using the known values:

  1. For \(n = 1\): \(A(1^2) + B(1) + C = 50\) → \(A + B + C = 50\)
  2. For \(n = 2\): \(A(2^2) + B(2) + C = -10\) → \(4A + 2B + C = -10\)
  3. For \(n = 3\): \(A(3^2) + B(3) + C = 2\) → \(9A + 3B + C = 2\)

Now solving the system of equations:

  1. \(A + B + C = 50\) (Equation 1)
  2. \(4A + 2B + C = -10\) (Equation 2)
  3. \(9A + 3B + C = 2\) (Equation 3)

Let's subtract Equation 1 from Equation 2: \[ (4A + 2B + C) - (A + B + C) = -10 - 50 \] \[ 3A + B = -60 \quad \text{(Equation 4)} \]

Now subtract Equation 2 from Equation 3: \[ (9A + 3B + C) - (4A + 2B + C) = 2 - (-10) \] \[ 5A + B = 12 \quad \text{(Equation 5)} \]

Now, we can eliminate \(B\) by subtracting Equation 4 from Equation 5: \[ (5A + B) - (3A + B) = 12 + 60 \] \[ 2A = 72 \implies A = 36 \]

Now substitute \(A = 36\) back into Equation 4: \[ 3(36) + B = -60 \] \[ 108 + B = -60 \implies B = -168 \]

Now substitute \(A\) and \(B\) back into Equation 1: \[ 36 - 168 + C = 50 \] \[ C = 50 + 168 - 36 = 182 \]

Now we have \(A\), \(B\), and \(C\):

  • \(A = 36\)
  • \(B = -168\)
  • \(C = 182\)

Thus, the formula for the nth term of the sequence is: \[ a_n = 36n^2 - 168n + 182 \]

You can verify this formula with \(n=1\), \(n=2\), and \(n=3\):

  • For \(n=1\): \(36(1^2) - 168(1) + 182 = 36 - 168 + 182 = 50\)
  • For \(n=2\): \(36(2^2) - 168(2) + 182 = 144 - 336 + 182 = -10\)
  • For \(n=3\): \(36(3^2) - 168(3) + 182 = 324 - 504 + 182 = 2\)

Thus, the nth term of the sequence is \(a_n = 36n^2 - 168n + 182\).