Question 1

AP with negative difference: 13, 8, 3, −2, … Find the 10th term.

Accepted Answer

Introduction / Context: This AP decreases by a constant amount each step. The nth-term formula still applies regardless of sign of the common difference. Given Data / Assumptions: a_1 = 13, d = −5 (since 8−13 = −5). Find a_10. Concept / Approach: Use a_n = a_1 + (n−1)d. Substitute n = 10, a_1 = 13, d = −5. Step-by-Step Solution: a_10 = 13 + 9*(−5) = 13 − 45 = −32. Verification / Alternative check: Compute a_5 = −2, then proceed five more steps of −5 each to get −27, −32 (consistent at a_10). Why Other Options Are Wrong: −64 and −30 are off the correct count of steps; −13 corresponds to a_6, not a_10. Common Pitfalls: Using n*d instead of (n−1)d, or miscounting negative increments. Final Answer: - 32

Question 2

AP identification: 10, 8, 6, 4, … What term of this AP is −28?

Accepted Answer

Introduction / Context: Solving “which term equals a specific value” in an AP means solving a linear equation in n using the nth-term form. Given Data / Assumptions: a_1 = 10, d = −2. Find n such that a_n = −28. Concept / Approach: Use a_n = a_1 + (n−1)d and solve for n. Step-by-Step Solution: a_n = 10 + (n−1)(−2) = 10 − 2n + 2 = 12 − 2n.Set 12 − 2n = −28 ⇒ −2n = −40 ⇒ n = 20. Verification / Alternative check: Compute a_20 directly: 10 + 19*(−2) = 10 − 38 = −28 (confirms). Why Other Options Are Wrong: 15, 18, 19 correspond to intermediate terms whose values are −18, −24, −26 respectively, not −28. Common Pitfalls: Algebraic slip when distributing the negative sign or using n instead of (n−1). Final Answer: 20

Question 3

AP constraints: The 9th term exceeds the 5th term by 32, and their sum is 114. Find the 8th term of the AP.

Accepted Answer

Introduction / Context: Two linear constraints on terms of an AP allow solving for the first term a and common difference d. Once a and d are known, any term is immediate. Given Data / Assumptions: t_n = a + (n−1)d. t_9 − t_5 = 32; t_9 + t_5 = 114. Concept / Approach: Write both constraints in a and d, solve the linear system, then compute t_8. Step-by-Step Solution: t_9 − t_5 = (a+8d) − (a+4d) = 4d = 32 ⇒ d = 8.t_9 + t_5 = (a+8d) + (a+4d) = 2a + 12d = 114 ⇒ 2a + 96 = 114 ⇒ a = 9.t_8 = a + 7d = 9 + 56 = 65. Verification / Alternative check: Compute t_5 = 9 + 4*8 = 41 and t_9 = 9 + 8*8 = 73; difference 32 and sum 114 confirm. Why Other Options Are Wrong: 60, 63, 68 are not equal to a + 7d with a = 9, d = 8. Common Pitfalls: Using n instead of (n−1) in t_n; arithmetic slips when adding the equations. Final Answer: 65

Question 4

AP recognition: 1, 8, 15, … (constant difference 7) Find the 11th term.

Accepted Answer

Introduction / Context: The sequence increases by 7 each time, a classic AP scenario solvable by the nth-term formula. Given Data / Assumptions: a_1 = 1, d = 7. Find a_11. Concept / Approach: Use a_n = a_1 + (n−1)d. Step-by-Step Solution: a_11 = 1 + 10*7 = 71. Verification / Alternative check: Compute a_6 = 36; five more steps of +7 yield 71 at the 11th term. Why Other Options Are Wrong: 69, 72, 73 are close but do not match the exact +7 progression to the 11th term. Common Pitfalls: Adding 7 eleven times instead of ten (since the first term already counts as n=1). Final Answer: 71

Question 5

Evaluate the alternating triple pattern: 1 − 2 − 3 + 2 − 3 − 4 + 3 − 4 − 5 + … up to 100 terms. Find the total sum.

Accepted Answer

Introduction / Context: This series groups naturally into blocks of three terms: +n, −(n+1), −(n+2). Summing by blocks streamlines evaluation of long finite series without writing every term. Given Data / Assumptions: Patterned blocks: for n = 1, 2, 3, … each block is n − (n+1) − (n+2). Total terms required: 100. Concept / Approach: First compute the sum per block S_n, then determine how many full 3-term blocks fit in 100, and add any leftover initial terms of the next block. Step-by-Step Solution: Block sum: S_n = n − (n+1) − (n+2) = −(n+3).In 100 terms there are 33 full blocks (99 terms) and 1 extra term.Sum of 33 blocks = −∑_{n=1}^{33} (n+3) = −( (33*34)/2 + 3*33 ) = −(561 + 99) = −660.The 100th term is the first term of the 34th block, i.e., +34.Total = −660 + 34 = −626. Verification / Alternative check: Spot-check early partial sums (after 3, 6, 9 terms) to confirm the −(n+3) block behavior before trusting the general calculation. Why Other Options Are Wrong: −624 and −676 reflect arithmetic slips (e.g., miscounting blocks or the final leftover term). −694 is too negative. Common Pitfalls: Forgetting the lone 100th term after 33 blocks; sign errors when combining positives and negatives in the block. Final Answer: -626

Question 6

Sum of all multiples of 8 between 200 and 400 (inclusive). Compute the arithmetic progression sum.

Accepted Answer

Introduction / Context: All integers divisible by a fixed number form an AP. Summing such terms over a closed interval reduces to counting how many terms fall in range and applying the AP sum formula. Given Data / Assumptions: First multiple ≥ 200: 200 (25*8). Last multiple ≤ 400: 400 (50*8). d = 8. Concept / Approach: Find n = ((last − first)/d) + 1, then S_n = n/2 * (first + last). Step-by-Step Solution: n = (400 − 200)/8 + 1 = 25 + 1 = 26.S = 26/2 * (200 + 400) = 13 * 600 = 7800. Verification / Alternative check: Average term is (200+400)/2 = 300; with 26 terms, 26*300 = 7800, consistent. Why Other Options Are Wrong: 7600, 7200, 7900 correspond to miscounting n or using wrong endpoints. Common Pitfalls: Accidentally excluding endpoints even though both are divisible by 8; using 24 or 25 terms instead of 26. Final Answer: 7800

Question 7

How many terms of the AP −20, −16, −12, … must be taken so that the sum equals 120?

Accepted Answer

Introduction / Context: Determining how many terms are needed to reach a target sum in an AP is a standard application of the partial-sum formula, producing a quadratic equation in n. Given Data / Assumptions: a_1 = −20, d = 4. Find n such that S_n = 120. Concept / Approach: Use S_n = n/2 * [2a_1 + (n−1)d], set equal to 120, and solve for n. Step-by-Step Solution: S_n = n/2 * (−40 + 4(n−1)) = n/2 * (4n − 44) = n(2n − 22).Set 2n^2 − 22n = 120 ⇒ 2n^2 − 22n − 120 = 0 ⇒ n^2 − 11n − 60 = 0.Solve: (n − 15)(n + 4) = 0 ⇒ n = 15 (positive solution). Verification / Alternative check: Compute S_15 quickly: average term * number of terms. a_15 = −20 + 14*4 = 36; average = (−20 + 36)/2 = 8; S_15 = 15 * 8 = 120. Why Other Options Are Wrong: 10, 20, 5 produce sums distinct from 120 (either too small or too large). Common Pitfalls: Dropping the factor 1/2; sign mistakes in 2a_1; forgetting S_n forms a quadratic in n. Final Answer: 15

Question 8

Find the sum of all even numbers up to and including 1672.

Accepted Answer

Introduction / Context: Even numbers form the AP: 2, 4, 6, …, 1672. Summing an AP is routine using count and average (or direct sum formula). Given Data / Assumptions: First term 2, last term 1672, d = 2. Include 1672 because the prompt says “including it”. Concept / Approach: Compute n = last/2, then S_n = n/2 * (first + last). Step-by-Step Solution: n = 1672 / 2 = 836.S = 836/2 * (2 + 1672) = 418 * 1674 = 699,732. Verification / Alternative check: Equivalently, S = n * average = 836 * (2 + 1672)/2 = 836 * 837 = 699,732. Why Other Options Are Wrong: 699,731 and 699,831/699,832 miss by ±1 or ±100 due to arithmetic slips. Common Pitfalls: Excluding the endpoint; miscomputing n or (first + last). Final Answer: 699732

Question 9

Find the sum of all even numbers less than or equal to 280 (i.e., below 281).

Accepted Answer

Introduction / Context: Even integers up to 280 form an AP with d = 2. Summation follows the standard AP method. Given Data / Assumptions: First term = 2, last term = 280, d = 2. “Below 281” implies including 280. Concept / Approach: Compute term count n = last/2, then apply S_n = n/2 * (first + last). Step-by-Step Solution: n = 280/2 = 140.S = 140/2 * (2 + 280) = 70 * 282 = 19,740. Verification / Alternative check: S = n * average = 140 * (2 + 280)/2 = 140 * 141 = 19,740. Why Other Options Are Wrong: 19730/19750/19760 are off by one step in averaging or by arithmetic slip. Common Pitfalls: Using 139 or 141 terms mistakenly; forgetting that 280 is even and should be included. Final Answer: 19740

Question 10

In an AP, the sum to 12 terms equals the sum to 18 terms. Find the sum to 30 terms of this AP.

Accepted Answer

Introduction / Context: Equating two partial sums in an AP creates a condition on a (first term) and d (common difference). Once a relation is found, we can deduce other sums efficiently. Given Data / Assumptions: S_n = n/2 * (2a + (n−1)d). S_12 = S_18. Goal: S_30. Concept / Approach: Set 12/2(2a+11d) = 18/2(2a+17d) ⇒ 6(2a+11d) = 9(2a+17d). Solve for a in terms of d; then compute S_30. Step-by-Step Solution: 12a + 66d = 18a + 153d ⇒ 0 = 6a + 87d ⇒ a = −87d/6 = −14.5d.S_30 = 30/2 * (2a + 29d) = 15(2(−14.5d) + 29d) = 15(−29d + 29d) = 0. Verification / Alternative check: If S_12 = S_18, then the sum from term 13 to 18 must be zero; symmetrical reasoning extends to S_30 = 0 under the derived a–d relation. Why Other Options Are Wrong: 1, 2, 3 contradict the established a–d relation; only 0 satisfies the identity for all d ≠ 0 implied by the condition. Common Pitfalls: Arithmetic mistakes when distributing 6 and 9; forgetting to halve when computing n/2. Final Answer: 0

Question 11

Population growth (doubling each hour): starting with 50 bacteria, how many are born during the 12th hour?

Accepted Answer

Introduction / Context: For exponential growth that doubles each hour, the population after n hours is P(n) = P(0)*2^n. The “born during the nth hour” equals the increment from hour n−1 to hour n. Given Data / Assumptions: Initial P(0) = 50. Doubling every hour: P(n) = 50*2^n. We seek births in the 12th hour: P(12) − P(11). Concept / Approach: Compute the increment as P(12) − P(11) = 50*2^12 − 50*2^11 = 50*2^11. Step-by-Step Solution: 2^11 = 2048.Births in 12th hour = 50 * 2048 = 102,400. Verification / Alternative check: Interpretation check: “born in the 12th hour” means increase during that hour, not total to date or population at the end. Why Other Options Are Wrong: 120,400 and 120,450 correspond to misapplied exponents; 102,460 is a rounding-like error (incorrect power or addition). Common Pitfalls: Using 2^12 instead of 2^11 for births; answering with total population at hour 12 rather than births during that hour. Final Answer: 102400

Question 12

Geometric earnings: ₹5 on day 1, ₹15 on day 2, ₹45 on day 3, and so on (ratio 3). How much does the boy earn in total by the end of 7 days?

Accepted Answer

Introduction / Context: The daily earnings form a geometric progression (GP) with first term a = 5 and common ratio r = 3. The total by day 7 is the sum of the first 7 GP terms. Given Data / Assumptions: a = 5 (₹), r = 3. Number of days n = 7. Concept / Approach: Use the GP sum formula for r ≠ 1: S_n = a * (r^n − 1) / (r − 1). Step-by-Step Solution: S_7 = 5 * (3^7 − 1) / (3 − 1) = 5 * (2187 − 1) / 2 = 5 * 2186 / 2 = 5 * 1093 = 5465. Verification / Alternative check: List last few terms: day 5 = 405, day 6 = 1215, day 7 = 3645; adding with prior days reaches 5465, matching the formula. Why Other Options Are Wrong: ₹ 5450, ₹ 5484, ₹ 5486 are near-misses typical of arithmetic slips or truncation; only ₹ 5465 is exact. Common Pitfalls: Confusing “amount on the 7th day” (which is 3645) with “total by 7 days” (which is 5465). Final Answer: ₹ 5465

Question 13

Ages – father as a linear combination of children: The father of two children has an age equal to twice the elder child’s age plus four times the younger child’s age. The geometric mean (GM) of the children’s ages is 4√3, and their harmonic mean (HM…

Accepted Answer

Introduction / Context: This is a number-theory/algebra blend using classical means (geometric and harmonic) to deduce two unknown ages. Once the children’s ages are identified, the father’s age is obtained from a linear combination: 2 × (elder) + 4 × (younger). Given Data / Assumptions: Let children’s ages be x (elder) and y (younger), x ≥ y. GM = √(xy) = 4√3 ⇒ xy = 48. HM = 2xy / (x + y) = 6 ⇒ x + y = 2xy / 6 = 16. Father’s age F = 2x + 4y. Concept / Approach: Use Vieta-like relationships from the quadratic with sum and product: t^2 − (x + y)t + xy = 0 ⇒ t^2 − 16t + 48 = 0. Solve to get the two ages, then compute F = 2x + 4y. Step-by-Step Solution: x + y = 16, xy = 48.t^2 − 16t + 48 = 0 ⇒ Discriminant = 256 − 192 = 64.t = (16 ± 8) / 2 ⇒ t ∈ {12, 4} ⇒ (elder, younger) = (12, 4).F = 2*12 + 4*4 = 24 + 16 = 40 yr. Verification / Alternative check: GM = √(12*4) = √48 = 4√3; HM = 2*12*4/(12+4) = 96/16 = 6 — both match, confirming ages 12 and 4 and hence F = 40 yr. Why Other Options Are Wrong: 32, 48, 56, 44 yr do not equal 2*12 + 4*4 and do not satisfy the derived constraints simultaneously. Common Pitfalls: Mixing up which mean formula to use; computing GM or HM incorrectly; reversing coefficients in F. Final Answer: 40 yr

Question 14

Geometric progression (GP) – find 4th term: In a GP, the 5th term is 81 and the first term is 16. Determine the 4th term.

Accepted Answer

Introduction / Context: Given two terms of a geometric progression (GP), we can find the common ratio and then any required term using the standard formula a_n = a * r^(n−1). Given Data / Assumptions: a (first term) = 16 a_5 = 81 GP term formula: a_n = a * r^(n−1) Concept / Approach: From a_5 = a * r^4, compute r^4 = 81/16. Recognize 81 = 3^4 and 16 = 2^4 ⇒ r = 3/2 (taking the positive ratio for a standard increasing GP). Then a_4 = a * r^3. Step-by-Step Solution: r^4 = 81/16 ⇒ r = 3/2.a_4 = 16 * (3/2)^3 = 16 * 27/8 = 2 * 27 = 54. Verification / Alternative check: a_5 = 16 * (3/2)^4 = 16 * 81/16 = 81, consistent with the given data. Why Other Options Are Wrong: 36, 24, 18, 48 arise from incorrect powers or mistaken ratio extraction. Common Pitfalls: Using r^5 instead of r^4; forgetting that the 4th term uses r^3, not r^4. Final Answer: 54

Question 15

Geometric progression – count terms up to a given term: How many terms are there in the GP 5, 20, 80, 320, …, 20480?

Accepted Answer

Introduction / Context: Counting terms in a GP reduces to solving a_n = a * r^(n−1) for n, given first term, common ratio, and last term. Given Data / Assumptions: a = 5 Common ratio r = 20/5 = 4 Last term L = 20480 Concept / Approach: Set 5 * 4^(n−1) = 20480 ⇒ 4^(n−1) = 4096. Express 4096 as a power of 4 to find n − 1. Step-by-Step Solution: 4^5 = 1024, 4^6 = 4096.Thus 4^(n−1) = 4^6 ⇒ n − 1 = 6 ⇒ n = 7. Verification / Alternative check: The sequence terms are 5, 20, 80, 320, 1280, 5120, 20480 — exactly 7 terms. Why Other Options Are Wrong: 6 or 5 underestimate; 8 or 10 overestimate the power index required to reach 20480. Common Pitfalls: Dividing by r incorrectly or misidentifying r from adjacent terms. Final Answer: 7

Question 16

Means – possible value with AM = 25 and GM = 7: Two positive numbers A and B have arithmetic mean 25 and geometric mean 7. Which of the following could be A?

Accepted Answer

Introduction / Context: Given AM and GM of two positive numbers, we can reconstruct them up to order by solving a quadratic with known sum and product. The question asks which listed value can be one of the numbers. Given Data / Assumptions: (A + B)/2 = 25 ⇒ A + B = 50 √(AB) = 7 ⇒ AB = 49 A, B > 0 Concept / Approach: Let t be a number satisfying t^2 − (A + B)t + AB = 0 ⇒ t^2 − 50t + 49 = 0. The roots are the possible values for A and B (order-free). Step-by-Step Solution: Discriminant Δ = 50^2 − 4*49 = 2500 − 196 = 2304.√Δ = 48 ⇒ t = (50 ± 48)/2 ⇒ t ∈ {49, 1}.Thus {A, B} = {49, 1}; one valid value for A is 49. Verification / Alternative check: AM = (49 + 1)/2 = 25, GM = √(49*1) = 7 — both conditions satisfied. Why Other Options Are Wrong: 10, 20, 25 do not complete a pair with sum 50 and product 49; 1 would also be valid but is not given in the original options set for selection of A (included here only to clarify the pair). Common Pitfalls: Misusing AM/GM inequalities or assuming A = B (which would give 25, not 7 as GM). Final Answer: 49

Question 17

Harmonic progression (HP) – find the 10th term: The second term of an HP is 1/6 and the fourth term is 1/14. Find the 10th term of the HP.

Accepted Answer

Introduction / Context: In a harmonic progression (HP), the reciprocals of the terms form an arithmetic progression (AP). Use this transformation to compute the required term index-wise, then invert to return to HP. Given Data / Assumptions: HP terms: H2 = 1/6, H4 = 1/14. Define AP terms A_n = 1/H_n. Thus A2 = 6 and A4 = 14. Concept / Approach: Let A_n = A + (n − 1)d. Using A2 and A4, solve for A and d; compute A10, then take H10 = 1/A10. Step-by-Step Solution: A + d = 6 … (1)A + 3d = 14 … (2)Subtract (2) − (1): 2d = 8 ⇒ d = 4.A = 6 − d = 2.A10 = A + 9d = 2 + 36 = 38 ⇒ H10 = 1/38. Verification / Alternative check: Check indices: A2 = 2 + 4 = 6 (OK); A4 = 2 + 12 = 14 (OK). Thus inversion back to HP is consistent. Why Other Options Are Wrong: 1/32, 1/36, 1/40, 1/30 correspond to different A10 values than the derived 38. Common Pitfalls: Forgetting to invert at the end; using n in place of (n − 1) inside the AP term formula. Final Answer: 1/38

Question 18

Arithmetic progression (AP) – compute the 12th term: If five times the fifth term of an AP equals seven times the seventh term, determine the twelfth term.

Accepted Answer

Introduction / Context: Relating two AP terms gives a linear equation in the first term and common difference. Once the relation is solved, the requested term follows by substitution. Given Data / Assumptions: a_n = a + (n − 1)d. 5(a + 4d) = 7(a + 6d). Concept / Approach: Expand and collect terms to express a in terms of d, then compute a_12 = a + 11d. Step-by-Step Solution: 5a + 20d = 7a + 42d ⇒ 0 = 2a + 22d ⇒ a = −11d.a_12 = a + 11d = (−11d) + 11d = 0. Verification / Alternative check: If a = −11 and d = 1, then a_5 = −7; a_7 = −5; 5(−7) = −35; 7(−5) = −35 (condition holds), and a_12 = 0. Why Other Options Are Wrong: Nonzero values contradict the derived identity a = −11d which forces a_12 = 0 for all d. Common Pitfalls: Plugging n instead of n − 1; arithmetic slips when moving terms across the equality. Final Answer: 0

Question 19

Number series – insert the missing term: 2, 8, 18, ?, 50, 72

Accepted Answer

Introduction / Context: This is a pattern-recognition series. Observing values against simple polynomial forms (squares, multiples) often reveals the rule efficiently. Observation / Approach: Check 2n^2: for n = 1..6 we get 2, 8, 18, 32, 50, 72 — exactly the given positions with one missing term. Thus a_n = 2n^2 fits perfectly. Step-by-Step Solution: n=1 → 2; n=2 → 8; n=3 → 18; n=4 → 32; n=5 → 50; n=6 → 72.The missing fourth term is 32. Verification / Alternative check: Second differences are constant: first differences 6, 10, 14, 18, 22; second differences all 4 — confirming a quadratic with leading coefficient 2. Why Other Options Are Wrong: 33, 36, 38, 30 break the exact 2n^2 rule and disrupt constant second differences. Common Pitfalls: Overfitting with ad-hoc difference jumps instead of recognizing a simple quadratic pattern. Final Answer: 32

Question 20

Number series – continue the cube-plus pattern: 2, 9, 28, 65, 126,  …  What is the next term?

Accepted Answer

Introduction / Context: Many competition series come from simple algebraic forms (squares or cubes) with small adjustments. Here, the values align with n^3 + 1. Observation / Approach: 1^3 + 1 = 2 2^3 + 1 = 9 3^3 + 1 = 28 4^3 + 1 = 65 5^3 + 1 = 126 Step-by-Step Solution: Next uses n = 6: 6^3 + 1 = 216 + 1 = 217. Verification / Alternative check: Differences grow as 7, 19, 37, 61, which match the expected cubic growth when adding 1 to n^3. Why Other Options Are Wrong: 195, 199, 205, 208 do not equal 6^3 + 1; they would break the clean cubic rule. Common Pitfalls: Trying to fit differences to ad-hoc multipliers rather than spotting the direct polynomial pattern. Final Answer: 217

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