Question

20, 30, 42, 56, 72, ____

• A. 82
• B. 90
• C. 92
• D. 94

Hint:

If first differences form their own simple AP, extend that AP rather than guessing the next term directly.

Answer 1

The Correct Option is B

Step 1: Compute differences
$30-20=10,\; 42-30=12,\; 56-42=14,\; 72-56=16$.
The increments are $10,12,14,16$—an arithmetic progression with common difference $2$.
Step 2: Extend the AP of differences
Next increment $=16+2=18$.
Step 3: Add to the last term
$72+18=90$.
Verification: If we continued, the next would be $90+20=110$, preserving the pattern $10,12,14,16,18,20,\dots$
Why other options fail:
$82, 92, 94$ break the “+2 to each gap” rule at some point; only $90$ keeps the uniform step growth.

Answer 2

Instead of tracking the differences between terms, check whether each number is a product of two consecutive integers. Testing 20: \(4 \times 5 = 20\). Testing 30: \(5 \times 6 = 30\). Testing 42: \(6 \times 7 = 42\). Testing 56: \(7 \times 8 = 56\). Testing 72: \(8 \times 9 = 72\). Every term fits the pattern of two consecutive integers multiplied together, with the smaller integer increasing by 1 each time: 4, 5, 6, 7, 8.

For the sixth term, the smaller integer continues the count to 9, so the value should be \(9 \times 10 = 90\).

1. 82: This is not the product of two consecutive integers near 9 and 10; \(9 \times 10 = 90\), not 82, so this breaks the pattern.
2. 90: This equals \(9 \times 10\), exactly continuing the consecutive-integer-product pattern that every earlier term followed.
3. 92: This is 2 more than 90 and does not equal the product of any pair of consecutive integers around this range, so it does not fit the pattern.
4. 94: This is 4 more than 90 and, like 92, is not a product of two consecutive integers, so it also breaks the established rule.

Since every earlier term is the product of two consecutive integers increasing by one each time, the sixth term must be \(9 \times 10 = 90\).

Therefore, the correct answer is 90.

Answer 3

To find a formula for this sequence rigorously rather than by testing guesses, model each term as a quadratic in its position, \( T_n = An^2 + Bn + C \), and solve for the constants using the first three terms.

At \(n=1\): \(A + B + C = 20\). At \(n=2\): \(4A + 2B + C = 30\). At \(n=3\): \(9A + 3B + C = 42\). Subtracting the first from the second gives \(3A + B = 10\). Subtracting the second from the third gives \(5A + B = 12\). Subtracting these two results gives \(2A = 2\), so \(A = 1\). Then \(B = 10 - 3 = 7\), and from the first equation, \(C = 20 - 1 - 7 = 12\).

So the model is \( T_n = n^2 + 7n + 12 \). Checking against the fourth and fifth terms confirms it: at \(n=4\), \(16 + 28 + 12 = 56\), matching, and at \(n=5\), \(25 + 35 + 12 = 72\), matching as well. For the sixth term, \(n=6\): \(T_6 = 36 + 42 + 12 = 90\).

1. 82: Substituting \(n=6\) into \(T_n = n^2+7n+12\) gives 90, not 82, so this does not satisfy the fitted formula.
2. 90: This exactly matches \(T_6 = 36+42+12=90\), confirming the quadratic model derived and verified against every known term.
3. 92: This is 2 more than the model's prediction of 90 and does not solve \(n^2+7n+12=92\) at \(n=6\).
4. 94: This overshoots the model's prediction by 4 and does not fit the formula at \(n=6\) either.

Solving the quadratic model algebraically and verifying it against every known term confirms the sixth term is 90.

Therefore, the correct answer is 90.