UGC NET Questions (Paper – 1)

This sequence can be modelled by aₙ = 2n⁴ + n! with n starting from 1. For n = 1, 2, 3, 4 and 5 we obtain 2·1⁴ + 1! = 3, 2·2⁴ + 2! = 34, 2·3⁴ + 3! = 168, 2·4⁴ + 4! = 536 and 2·5⁴ + 5! = 1370, which matches the given terms. For n = 6 we have a₆ = 2·6⁴ + 6! = 2·1296 + 720 = 2592 + 720 = 3312. Hence 3312 is the unique value that maintains this factorial-quartic pattern.

Option A:
Option A, 3288, is 24 less than the value produced by 2n⁴ + n! at n = 6. Choosing 3288 would mean reducing the correct outcome of the formula only at the final term, which is not supported by earlier behaviour. Therefore option A is not consistent with the established rule.

Option B:
Option B, 3300, is 12 smaller than the correct value and cannot be written as 2·1296 + 720. It is numerically close but does not respect the exact functional dependence on n. As a result, option B does not correctly continue the series.

Option C:
Option C, 3324, is 12 greater than the correct term and again cannot be obtained from 2·6⁴ + 6!. Accepting 3324 would force an unjustified increase in the last value relative to the pattern. Thus option C is not a valid continuation.

Option D:
Option D, 3312, matches exactly the number generated by the rule aₙ = 2n⁴ + n! for n = 6. It preserves the relationship that combines a strong quartic component with factorial growth evident in previous terms. Because it extends the same pattern smoothly, this option is the correct answer.

The pattern can be described by aₙ = 2n! + n² with n starting from 1. For n = 1, 2, 3, 4 and 5, this yields 2·1! + 1² = 3, 2·2! + 2² = 8, 2·3! + 3² = 21, 2·4! + 4² = 64 and 2·5! + 5² = 265, matching the given terms. For n = 6, we obtain a₆ = 2·6! + 6² = 2·720 + 36 = 1476. Thus 1476 is the only value that fits this factorial-based rule and continues the series.

Option A:
Option A, 1452, is 24 less than the correct value given by 2n! + n² at n = 6. To pick 1452 we would need to subtract 24 from the computed result only at this position, which would disrupt the pattern of combining factorial and square terms. Therefore option A is incorrect.

Option B:
Option B, 1476, exactly matches the number produced by 2·6! + 6². It preserves the factorial component and the quadratic addition in the same way as for earlier values of n. Because it extends the rule without any modification, this option correctly continues the series.

Option C:
Option C, 1464, is 12 smaller than the correct value and cannot be expressed as 2·720 + 36. It is only an approximate candidate and does not reflect the precise functional relationship. As such, option C is not a valid next term.

Option D:
Option D, 1488, is 12 greater than the required value and again fails to satisfy the formula at n = 6. Choosing 1488 would overstate the term in comparison to what the rule dictates, breaking the pattern. Hence option D is not acceptable.

This sequence is defined by a₁ = 3 and aₙ₊₁ = 4aₙ + n² + 1 for n ≥ 1. Applying the rule yields a₂ = 4·3 + 1² + 1 = 14, a₃ = 4·14 + 2² + 1 = 61, a₄ = 4·61 + 3² + 1 = 254 and a₅ = 4·254 + 4² + 1 = 1033, matching the given terms. For n = 5 the next term is a₆ = 4·1033 + 5² + 1 = 4158. Therefore 4158 is the unique number that continues the same recurrence correctly.

Option A:
Option A, 4126, is 32 less than the recurrence output for n = 5. To obtain 4126 we would need to subtract 32 from the formula result at the last step, which is not consistent with the earlier pattern. Hence this option does not follow the rule and is incorrect.

Option B:
Option B, 4142, is closer but still 16 less than the correct value 4158. It cannot be expressed as 4a₅ + 5² + 1 and would imply undercounting the contribution of the square term. As the recurrence is satisfied exactly before, option B cannot be the correct continuation.

Option C:
Option C, 4148, is slightly below the required value and also fails to satisfy the recurrence exactly. Selecting 4148 would again require modifying the output of the rule only at this step. Such a change destroys the strict structure of the sequence, so this option is not valid.

Option D:
Option D, 4158, equals exactly the result produced by aₙ₊₁ = 4aₙ + n² + 1 when n = 5. It maintains the process of quadrupling the previous term and adding a quadratic plus constant correction. Because this description explains each transition and leads naturally to 4158, option D is correct.

The sequence can be defined recursively by a₁ = 7 and aₙ₊₁ = 3aₙ + n² for n ≥ 1. Using this rule, a₂ = 3·7 + 1² = 22, a₃ = 3·22 + 2² = 70, a₄ = 3·70 + 3² = 219 and a₅ = 3·219 + 4² = 673, matching the given terms. For n = 5 the next term is a₆ = 3·673 + 5² = 2044. Thus 2044 is the only value that satisfies the same recurrence and correctly continues the series.

Option A:
Option A, 2044, is exactly the result produced by aₙ₊₁ = 3aₙ + n² when it is applied to a₅ with n = 5. It preserves the pattern of tripling the previous term and adding the square of the index. Because this rule explains each step of the sequence and yields 2044 next, this option is correct.

Option B:
Option B, 2032, is 12 less than the recurrence output for n = 5. To accept 2032 we would have to arbitrarily subtract 12 from the correctly computed value while keeping the rule unchanged before. Such a disturbance is not justified by the earlier progression, so option B is incorrect.

Option C:
Option C, 2056, is 12 more than the required value from the recurrence. Selecting 2056 would similarly require increasing the formula result only at this final step. That would break the precise relationship described by aₙ₊₁ = 3aₙ + n², making option C invalid.

Option D:
Option D, 2072, deviates even further from the recurrence value and cannot be written as 3·673 + 5². Using 2072 would disregard the established rule and destroy the exact fit observed in earlier terms. Therefore this option does not correctly continue the series.

This sequence is modelled by the rule aₙ = n! + 3n² with n starting from 1. For n = 1, 2, 3, 4 and 5 the expression gives 1! + 3·1² = 4, 2! + 3·2² = 14, 3! + 3·3² = 33, 4! + 3·4² = 72 and 5! + 3·5² = 195, which matches the given terms exactly. For n = 6 we obtain a₆ = 6! + 3·6² = 720 + 108 = 828. Hence 828 is the only value consistent with this factorial-based pattern.

Option A:
Option A, 804, is 24 less than the value from n! + 3n² at n = 6 and cannot be generated by the same formula. Choosing 804 would require subtracting 24 from the correct result only at the sixth term, which disrupts the pattern of factorial growth with a quadratic adjustment. Therefore option A is incorrect.

Option B:
Option B, 816, is 12 below the correct value 828 and does not equal 720 + 108. It represents a near approximation but still violates the exact functional relationship between n and aₙ. Consequently, option B does not correctly continue the series.

Option C:
Option C, 840, is 12 greater than the required value and likewise cannot arise from 6! + 3·6². Accepting 840 would artificially inflate the last term beyond what the rule predicts. Thus option C is not a valid continuation of the sequence.

Option D:
Option D, 828, exactly matches the value produced by the rule when n = 6. It faithfully maintains the combination of factorial growth and quadratic adjustment that governs all previous terms. Because it extends this pattern without any exception, option D is the correct answer.

This sequence is defined recursively by a₁ = 8 and aₙ₊₁ = 3aₙ + 2ⁿ + 1 for n ≥ 1. Using this rule gives a₂ = 3·8 + 2¹ + 1 = 27, a₃ = 3·27 + 2² + 1 = 86, a₄ = 3·86 + 2³ + 1 = 267 and a₅ = 3·267 + 2⁴ + 1 = 818, which exactly matches the given terms. For n = 5 the next term is a₆ = 3·818 + 2⁵ + 1 = 2454 + 32 + 1 = 2487. Hence 2487 is the unique value that satisfies this recurrence and continues the series correctly.

Option A:
Option A, 2463, is 24 less than the value obtained from 3a₅ + 2⁵ + 1 and cannot be generated by the recurrence. To get 2463 we would have to reduce the computed result only at this step, breaking the consistent update rule. Therefore option A is not correct.

Option B:
Option B, 2475, is 12 smaller than the correct term 2487 and again does not equal 3·818 + 32 + 1. It approximates the right value but does not follow from the recurrence formula. As such, option B cannot be considered a valid continuation of the sequence.

Option C:
Option C, 2487, exactly matches the value produced by the recurrence for n = 5. It maintains the structure of tripling the previous term and adding a power of 2 plus 1, which explains all prior transitions. Because it respects the rule fully, this option is the correct answer.

Option D:
Option D, 2499, is 12 greater than the correct value and would require increasing the additive part beyond 2⁵ + 1 at the final step. This change would alter the recurrence mechanism only at a single position and is not supported by earlier terms. Hence option D is incorrect.

The terms fit the polynomial rule aₙ = n⁴ + 3n³ + 2 with n starting from 1. Substituting n = 1, 2, 3, 4 and 5 into this expression yields 6, 42, 164, 450 and 1002, exactly the numbers in the series. Evaluating the same formula for n = 6 produces 1946. Hence 1946 is the only value that extends this quartic-plus-cubic pattern correctly.

Option A:
Option A, 1922, is 24 less than the value obtained from aₙ = n⁴ + 3n³ + 2 when n = 6. Accepting 1922 would require reducing the polynomial result only at the sixth term, which contradicts the behaviour of earlier entries. Therefore option A is not consistent with the established pattern.

Option B:
Option B, 1934, is still lower than the required value and does not equal the expression’s output at n = 6. It lies between the earlier predictions but fails to respect the exact algebraic rule. As a result, option B cannot be the correct continuation of the sequence.

Option C:
Option C, 1946, matches precisely the value produced by the formula when n = 6. It preserves the same contributions from the fourth-power and cubic terms together with the constant part. Because this rule explains all given terms and leads naturally to 1946, option C is the correct answer.

Option D:
Option D, 1962, is 16 greater than the polynomial value and cannot be generated by n⁴ + 3n³ + 2 at n = 6. Introducing such an extra increment would break the clean functional relationship. Hence option D is not acceptable.

This sequence follows the polynomial rule aₙ = 3n⁴ + n³ + n with n starting from 1. Evaluating the expression for n = 1, 2, 3, 4 and 5 yields 5, 58, 273, 836 and 2005, matching the given terms exactly. For n = 6, we obtain a₆ = 3·6⁴ + 6³ + 6 = 3·1296 + 216 + 6 = 3888 + 222 = 4110. Therefore 4110 is the only number that fits this functional pattern and correctly continues the series.

Option A:
Option A, 4086, is 24 less than the value predicted by 3n⁴ + n³ + n when n = 6. Choosing 4086 would require reducing the correct polynomial result at this term, even though earlier values are exact. This violates the consistency of the pattern, so option A is not acceptable.

Option B:
Option B, 4098, is 12 less than the correct value and again cannot be expressed as 3·6⁴ + 6³ + 6. It provides an approximation but not the true outcome of the formula. In a high-level series problem, such a discrepancy means the option does not represent the genuine rule.

Option C:
Option C, 4122, is 12 greater than the correct term 4110 and likewise does not match the polynomial output. Accepting 4122 would introduce an unexplained upward shift at the last step. Therefore this choice cannot be considered a valid continuation of the series.

Option D:
Option D, 4110, is exactly the value obtained when n = 6 is substituted into 3n⁴ + n³ + n. It preserves the same combination of a strong quartic term, a cubic correction and a linear part that explains the earlier numbers. Because it extends the rule seamlessly, option D is the correct answer.

The series is generated by the polynomial rule aₙ = n⁴ + 6n with n starting from 1. Substituting n = 1, 2, 3, 4 and 5 gives 7, 28, 99, 280 and 655, which exactly matches the given terms. Applying the same expression for n = 6 gives a₆ = 6⁴ + 6·6 = 1296 + 36 = 1332. Therefore 1332 is the only value that continues this polynomial pattern without changing the underlying rule.

Option A:
Option A gives 1332, which is exactly the value obtained from the formula n⁴ + 6n for n = 6. It maintains the same relationship between the index and term that produced all earlier numbers. Because no adjustment or exception is needed at the sixth term, this option correctly extends the series.

Option B:
Option B, 1308, is 24 less than the value predicted by n⁴ + 6n at n = 6. To accept 1308 we would have to reduce the correct polynomial result only at the last step, which contradicts the exact agreement seen for the first five terms. Hence this option does not respect the established pattern.

Option C:
Option C, 1320, is still below the correct value and cannot be obtained from n⁴ + 6n when n = 6. It represents a near but inaccurate approximation that breaks the strict algebraic rule behind the sequence. For a UGC NET numerical reasoning item, such a deviation means this option must be rejected.

Option D:
Option D, 1344, is 12 greater than the formula’s output and likewise cannot be produced by the same rule at n = 6. Choosing 1344 would introduce an unexplained increase in the last term, destroying the consistency of the pattern. Therefore this option is not a valid continuation of the number series.

The terms follow the rule aₙ = n⁴ + 5n with n starting from 1. Substituting n = 1, 2, 3, 4 and 5 gives 6, 26, 96, 276 and 650, exactly the given sequence. Applying the same expression for n = 6 produces 1326. Therefore 1326 is the unique value that continues this polynomial pattern without any change in the rule.

Option A:
Option A gives 1326, which is precisely the value obtained from aₙ = n⁴ + 5n when n = 6. It preserves the same functional relationship between n and aₙ that generates all earlier terms. Hence this option correctly represents the next term of the series.

Option B:
Option B proposes 1314, which is close but does not equal the value from n⁴ + 5n at n = 6. To obtain 1314 we would need to subtract 12 from the polynomial result only at this step. Since no such irregular adjustment appears in the previous terms, 1314 cannot be the correct continuation.

Option C:
Option C suggests 1338 as the next term. This number is 12 greater than the value predicted by the rule for n = 6. Introducing such an extra increment only at the sixth term would break the consistent algebraic pattern, so 1338 is not acceptable.

Option D:
Option D gives 1350, which deviates even further from the calculated value 1326. No single formula of the form n⁴ + 5n can generate the first five terms and then jump to 1350 next. Because it would require altering the rule at the last term, this option is incorrect.