Question:

53, 50, 45, 38, 29, ____

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When numbers fall quickly but not linearly, try “subtract consecutive odds/evens” or “add consecutive squares/cubes” before more complex rules.
Updated On: Jul 15, 2026
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The Correct Option is C

Approach Solution - 1

Step 1: Compute first differences
$50-53=-3,\; 45-50=-5,\; 38-45=-7,\; 29-38=-9$.
The sequence is strictly decreasing; the absolute differences are $3,5,7,9$—consecutive odd numbers.
Step 2: Extend the difference pattern
If the rule is “subtract consecutive odd numbers”, the next subtraction is $11$.
Step 3: Find the missing term and verify
$29-11=18$.
Check backward: $18$ would lead to the next term $18-(\text{next odd }13)=5$, which maintains the rule if continued. Hence the pattern is consistent.
Common pitfalls: Mistaking it for an arithmetic progression with constant difference, or for alternating operations. Always list consecutive gaps first.
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Approach Solution -2

Instead of tracking the differences between terms, check whether each number can be written as a constant minus a perfect square. Testing 53: \(54 - 1^2 = 54 - 1 = 53\). Testing 50: \(54 - 2^2 = 54 - 4 = 50\). Testing 45: \(54 - 3^2 = 54 - 9 = 45\). Testing 38: \(54 - 4^2 = 54 - 16 = 38\). Testing 29: \(54 - 5^2 = 54 - 25 = 29\). Every term fits the formula \(54 - n^2\), where \(n\) is the term's position in the sequence.

For the sixth term, \(n = 6\), so the value should be \(54 - 6^2 = 54 - 36 = 18\).

  1. 11: This does not equal \(54 - 36\); it is 7 less than what the formula predicts, so it breaks the pattern that fit all five earlier terms perfectly.
  2. 17: This is 1 less than 18, close but not exact, and does not equal \(54 - 6^2\), so it does not satisfy the formula either.
  3. 18: This equals \(54 - 6^2 = 54 - 36\), exactly continuing the constant-minus-square pattern that every earlier term followed without exception.
  4. 21: This overshoots the predicted value by 3 and does not fit \(54 - n^2\) for any small integer \(n\) near 6, so it breaks the established pattern.

Since every earlier term equals 54 minus the square of its position, the sixth term must equal \(54 - 36 = 18\).

Therefore, the correct answer is 18.

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Approach Solution -3

Rather than guessing that each term might be a constant minus a perfect square, we can derive this rigorously by modelling the sequence as a quadratic in its position, \( T_n = An^2 + Bn + C \), and solving for A, B and C using three known terms, then checking the fit against the rest.

Using the first three terms: at \(n=1\), \(A + B + C = 53\); at \(n=2\), \(4A + 2B + C = 50\); at \(n=3\), \(9A + 3B + C = 45\). Subtracting the first equation from the second gives \(3A + B = -3\). Subtracting the second from the third gives \(5A + B = -5\). Subtracting these two results gives \(2A = -2\), so \(A = -1\). Substituting back, \(3(-1) + B = -3\), so \(B = 0\). Then from the first equation, \(C = 53 - A - B = 53 + 1 - 0 = 54\).

So the model is \( T_n = -n^2 + 54 = 54 - n^2 \). Checking it against the fourth and fifth terms confirms the fit: at \(n=4\), \(54 - 16 = 38\), matching, and at \(n=5\), \(54 - 25 = 29\), matching as well. For the sixth term, \(n=6\): \(T_6 = 54 - 36 = 18\).

  1. 11: Substituting \(n=6\) into \(T_n = 54 - n^2\) gives 18, not 11, so this value does not satisfy the fitted quadratic model.
  2. 17: This is 1 short of the model's prediction of 18 at \(n=6\), so it does not solve \(54 - n^2 = 17\) for \(n=6\).
  3. 18: This exactly matches \(T_6 = 54 - 36 = 18\), confirming the quadratic model derived and verified against all five known terms.
  4. 21: This overshoots the model's prediction by 3 and does not fit \(54 - n^2\) at \(n=6\).

Solving the quadratic model algebraically and verifying it against every known term confirms the sixth term is 18.

Therefore, the correct answer is 18.

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