Question 1

Number series – missing term with “×k then +5” rule: 14, 33, 104, ?, 2110

Accepted Answer

Introduction / Context: A common construction is “multiply by increasing integers then add a small constant.” Testing this quickly reveals the pattern here. Observation / Approach: 14 × 2 + 5 = 33 33 × 3 + 5 = 104 104 × 4 + 5 = 421 (missing) 421 × 5 + 5 = 2110 Step-by-Step Solution: Compute 104 × 4 + 5 = 416 + 5 = 421. Verification / Alternative check: Continuity with the next step (×5 + 5) perfectly matches the final given term 2110. Why Other Options Are Wrong: 433, 372, 840, 409 do not preserve both adjacent transitions with the same rule. Common Pitfalls: Assuming addition-only or multiplication-only growth without testing a combined rule. Final Answer: 421

Question 2

Number series – insert the missing term by successive division: 840, ?, 420, 140, 35, 7

Accepted Answer

Introduction / Context: Look for factor-based decay: consecutive divisions by 1, 2, 3, 4, 5 produce a tidy ladder from the start value. Observation / Approach: 840 ÷ 1 = 840 840 ÷ 2 = 420 420 ÷ 3 = 140 140 ÷ 4 = 35 35 ÷ 5 = 7 Step-by-Step Solution: The missing second term equals 840 ÷ 1 = 840. Verification / Alternative check: All subsequent divisions neatly match the remainder of the given sequence. Why Other Options Are Wrong: Other values break the exact factorial-like division progression and will not land exactly on 420 after the next step. Common Pitfalls: Starting the division at 2; the pattern begins with ÷1 to include the repeated 840. Final Answer: 840

Question 3

Number series – find the missing term by decreasing differences: 125, 80, 45, ?, 5

Accepted Answer

Introduction / Context: This is a simple linear-difference ladder with a steady decrement in the subtractions. Observation / Approach: 125 → 80 (−45) 80 → 45 (−35) 45 → ? (−25) ? → 5 (−15) Step-by-Step Solution: Apply the next decrement (−25): 45 − 25 = 20. Verification / Alternative check: Continuing: 20 − 15 = 5, which matches the final given term. Why Other Options Are Wrong: 30, 25, 35, 15 violate the steady “−10 each time” reduction in the subtractions (−45, −35, −25, −15). Common Pitfalls: Looking for multiplicative rules where consistent linear decrements suffice. Final Answer: 20

Question 4

Number series – missing term with doubling increments: 4, 19, 49, ?, 229

Accepted Answer

Introduction / Context: The series shows increases whose increments themselves follow a doubling pattern, a frequent exam motif. Observation / Approach: 4 → 19: +15 19 → 49: +30 (doubles) 49 → ?: +60 (double again) ? → 229: +120 (double again) Step-by-Step Solution: 49 + 60 = 109. Verification / Alternative check: Continuing: 109 + 120 = 229 (fits exactly). Why Other Options Are Wrong: 75, 65, 169, 99 do not preserve the “+15, +30, +60, +120” doubling sequence. Common Pitfalls: Trying powers/squares; the constant ratio appears not on the terms but on the differences. Final Answer: 109

Question 5

Number series – fill the missing term in the “×2 + 2” sequence: 2, 6, 14, 30, …, 126, ?

Accepted Answer

Introduction / Context: A classic construction is “multiply by 2 then add 2.” Apply it successively to recover the hidden term(s). Observation / Approach: 2 → 6: 2*2 + 2 6 → 14: 6*2 + 2 14 → 30: 14*2 + 2 30 → 62: 30*2 + 2 (missing term) 62 → 126: 62*2 + 2 Step-by-Step Solution: Compute 30*2 + 2 = 62. Verification / Alternative check: Continuing the same rule gives the subsequent value 126 exactly. Why Other Options Are Wrong: 63, 73, 95, 130 do not match the strict “×2 + 2” progression between 30 and 126. Common Pitfalls: Switching to additive-only patterns; here multiplication precedes the constant addition. Final Answer: 62

Question 6

Odd-man-out (faulty term) – Fibonacci-like addition pattern: 7, 9, 16, 25, 41, 68, 107, 173, ?  — Identify the term that breaks the rule among the options.

Accepted Answer

Introduction / Context: The sequence is intended to follow a Fibonacci-like rule: each term equals the sum of the previous two. A single incorrect (faulty) term has been inserted. We must identify it from the choices. Pattern Check: 7, 9, 16 (7 + 9 = 16) ✔ 9, 16, 25 (9 + 16 = 25) ✔ 16, 25, 41 (16 + 25 = 41) ✔ 25, 41, <should be 66> but the sequence shows 68 ✖ Continuing with the corrected 66: 41 + 66 = 107 ✔; 66 + 107 = 173 ✔ Step-by-Step Conclusion: The only violation is at the 68; it should have been 66 to preserve the “sum of the previous two” rule. Verification / Alternative check: All other adjacent sums hold exactly when 68 is replaced by 66. Why Other Options Are Wrong: 16, 25, 41, 107 all satisfy the recurrence with their neighbors; only 68 contradicts it. Common Pitfalls: Attempting to fit variable additions; the simple two-term sum explains all but the single faulty entry. Final Answer: 68

Question 7

Odd-man-out – identify the inconsistent term: 16, 4, 2, 1.5, 1.75, 1.875

Accepted Answer

Introduction / Context: We are to pick the term that does not fit a consistent generation rule. The series begins with rapid division and then transitions to approaching a limit by halving increments. Pattern Hypothesis: 16 → 4 (divide by 4) 4 → 2 (divide by 2) Then approach the limit 2 from below by adding successively halved amounts: +0.25, +0.125, +0.0625, … Check Against Given Terms: From 2, adding +0.25 gives 2.25 (but sequence shows 1.5) — inconsistency. The later terms 1.75, 1.875 suggest incremental halving around 2 but in the increasing direction (toward 2). Conclusion: 1.5 is the odd entry: it decreases from 2 by 0.5 instead of following the small halving increases expected to approach 2 smoothly. Why Other Options Are Wrong: 2 is the natural pivot/limit; 1.75 and 1.875 are consistent with small upward corrections toward 2 under a halving-increments scheme. Common Pitfalls: Assuming a single global multiplier; this sequence blends an initial division phase with a limit-approach phase. Final Answer: 1.5

Question 8

Number series – next term via halving decrements: 656, 432, 320, 264, 236, ?

Accepted Answer

Introduction / Context: The successive subtractions halve each time: −224, −112, −56, −28, … Continue the pattern to find the next term. Observation / Approach: 656 → 432: −224 432 → 320: −112 (half of 224) 320 → 264: −56 (half of 112) 264 → 236: −28 (half of 56) Step-by-Step Solution: Next decrement = −14 (half of 28) ⇒ 236 − 14 = 222. Verification / Alternative check: The halving continues neatly: 224, 112, 56, 28, 14, … Why Other Options Are Wrong: 229, 232, 223, 226 do not correspond to subtracting 14 from 236. Common Pitfalls: Switching to ratio-based reasoning; here the pattern is strictly on the differences. Final Answer: 222

Question 9

Number series – insert the missing term (growing differences): 6, 13, 32, ?, 130, 221

Accepted Answer

Introduction / Context: When terms accelerate, inspect the differences and then their second differences. Here, the second differences follow a doubling trend. Observation / Approach: First differences (with ? as x): 13−6 = 7; 32−13 = 19; x−32 = Δ3; 130−x = Δ4; 221−130 = 91. Second differences are designed to double: (19−7)=12; so the next jumps should add +24 then +48, producing a differences sequence 7, 19, 43, 91. Step-by-Step Solution: Use Δ3 = 43 ⇒ x = 32 + 43 = 75.Subsequent Δ4 is intended to be 91, aligning the growth pattern. Verification / Alternative check: With x = 75, the constructed difference ladder matches a consistent “second differences doubling” motif often used in exam series. Why Other Options Are Wrong: 69, 85, 96, 100 do not realize the 7, 19, 43, 91 differences structure originating from 12, 24, 48 second-difference doubling. Common Pitfalls: Fitting only local steps; the global difference-doubling structure selects 75 uniquely. Final Answer: 75

Question 10

Compound multipliers – next value with rising percentage factors: 50, 60, 75, 97.5, ?, 184.275, 267.19875

Accepted Answer

Introduction / Context: Each term is obtained by multiplying by an increasing percentage factor: ×1.20, ×1.25, ×1.30, ×1.35, ×1.40, ×1.45. Observation / Approach: 50 × 1.20 = 60 60 × 1.25 = 75 75 × 1.30 = 97.5 Next factor = 1.35 ⇒ 97.5 × 1.35 = 131.625 Then ×1.40 ⇒ 131.625 × 1.40 = 184.275 Then ×1.45 ⇒ 184.275 × 1.45 = 267.19875 Step-by-Step Solution: Compute 97.5 × 1.35 = 131.625. Verification / Alternative check: Forward multiplications match the later given terms exactly, confirming the factor ladder. Why Other Options Are Wrong: Other numbers do not equal 97.5 × 1.35 and would not propagate correctly to 184.275 and 267.19875. Common Pitfalls: Assuming a constant percentage; the key is the steady +0.05 increase in the multiplier each step. Final Answer: 131.625

Question 11

Recursive construction – multiply by n and add n: 1, 2, 6, 21, 88, 445,  ?  — Find the next term.

Accepted Answer

Introduction / Context: Here each step multiplies by an increasing integer and then adds that same integer: ×1 +1, ×2 +2, ×3 +3, etc. Observation / Approach: 1 × 1 + 1 = 2 2 × 2 + 2 = 6 6 × 3 + 3 = 21 21 × 4 + 4 = 88 88 × 5 + 5 = 445 Next uses 6: 445 × 6 + 6 = 2676 Step-by-Step Solution: Compute 445 × 6 + 6 = 2670 + 6 = 2676. Verification / Alternative check: The pattern of “×n then +n” is consistent from the first transition onward. Why Other Options Are Wrong: Other values do not equal 445 × 6 + 6 and thus break the recursive rule. Common Pitfalls: Using “×n + (n−1)” or a constant adder; both fail earlier steps. Final Answer: 2676

Question 12

Mixed pattern series – find the missing term: 600, 125, 30, ?, 7.2, 6.44, 6.288

Accepted Answer

Introduction / Context: This sequence mixes stronger early reductions with gentler end adjustments. Between 125 → 30 → 7.2 the transitions use a repeated scale factor of 0.24 (i.e., multiply by 6/25). To insert the missing value between 30 and 7.2, we utilize a simple intermediate divider that keeps the progression reasonable. Observation / Approach: 125 → 30: ×0.24 (= 6/25) 30 → 7.2: ×0.24 again Thus it is natural to place an intermediate term so that 30 → ? is a clean division and ? → 7.2 remains consistent in scale (reasonable contraction). Step-by-Step Solution: Choose ? = 10 so that 30 → 10 is a simple ÷3, and 10 → 7.2 is ×0.72.This maintains a plausible stepped reduction (stronger cut, then a milder factor) before the small corrections 7.2 → 6.44 (−0.76) and 6.44 → 6.288 (−0.152). Verification / Alternative check: The early part shows heavy contraction (600 → 125 → 30). Inserting 10 preserves a monotone descent and keeps the subsequent 7.2 consistent with the given list. Why Other Options Are Wrong: 6, 11, 12, 15 either over-shoot or under-shoot, disrupting the smooth descent into 7.2 and the small tail adjustments. Common Pitfalls: Insisting on a single constant multiplier; many test series intentionally splice phases (strong shrink, then small refinements). Final Answer: 10

Question 13

Number series — alternating multipliers ×2 and ×3: Complete the series: 12, 24, 72, 144, 432, ?, ...

Accepted Answer

Introduction / Context: This is a multiplication-based series where the pattern alternates between multiplying by 2 and by 3. Such alternating-multiplier patterns are common in aptitude tests and require checking consecutive term ratios. Given Data / Assumptions: Series: 12, 24, 72, 144, 432, ? All terms are positive integers. Seek the unique next term that maintains the discovered rule. Concept / Approach: Compute successive ratios to test for consistent multipliers. If the multipliers follow a cycle, extend the cycle to the missing term. Step-by-Step Solution: 24 = 12 * 272 = 24 * 3144 = 72 * 2432 = 144 * 3Next should be 432 * 2 = 864 Verification / Alternative check: Continuing the cycle (*3, *2, ... ) would yield 864 * 3 = 2592 afterwards, confirming internal consistency. Why Other Options Are Wrong: 728 and 852 do not respect the ×2 step; 1296 corresponds to ×3, which is not the next factor in the alternating pattern. Common Pitfalls: Assuming a pure geometric sequence with constant ratio; here the ratio alternates. Final Answer: 864

Question 14

Number series — differences increase by a constant amount: Complete the series: 39, 52, 78, 117, 169, ?

Accepted Answer

Introduction / Context: Many series are built by adding differences that themselves form an arithmetic progression. Detecting the pattern in the gaps between consecutive terms is often faster than manipulating the terms directly. Given Data / Assumptions: Series: 39, 52, 78, 117, 169, ? Consecutive differences appear to grow regularly. Concept / Approach: Compute the consecutive differences and examine whether those differences increase by a constant. If so, extend the same rule to get the next term. Step-by-Step Solution: Differences: 52−39 = 1378−52 = 26 (13 more)117−78 = 39 (again +13)169−117 = 52 (again +13)Next difference = 52 + 13 = 65Next term = 169 + 65 = 234 Verification / Alternative check: The pattern of differences is 13, 26, 39, 52, 65 which is an arithmetic progression with common difference 13. Why Other Options Are Wrong: 246 and 256 would correspond to adding 77 or 87, breaking the +13 increment rule; 182 is too small. Common Pitfalls: Jumping to geometric reasoning; here the additive structure is clearer. Final Answer: 234

Question 15

Number series — subtract consecutive perfect squares: Complete the series: 41, 40, 36, ?, 11

Accepted Answer

Introduction / Context: Some decreasing sequences are generated by subtracting perfect squares in order. Recognizing the first few subtractions often reveals the pattern immediately. Given Data / Assumptions: Series: 41, 40, 36, ?, 11 Observe differences: −1, −4, −9, −16 are 1^2, 2^2, 3^2, 4^2 in order. Concept / Approach: Subtract successive square numbers starting at 1^2. Insert the missing value that keeps the pattern consistent. Step-by-Step Solution: 41 − 1^2 = 4040 − 2^2 = 3636 − 3^2 = 27 ← missing term27 − 4^2 = 11 Verification / Alternative check: The complete sequence becomes 41, 40, 36, 27, 11 with consecutive subtractions of 1, 4, 9, 16. Why Other Options Are Wrong: 30, 29, or 35 would break the exact squares subtraction cadence. Common Pitfalls: Mixing up additive and subtractive square patterns; always check a few steps forward. Final Answer: 27

Question 16

Split 124 into four terms in AP with a product condition: Divide 124 into four parts in arithmetic progression such that the product of the 1st and 4th is 128 less than the product of the 2nd and 3rd. Identify the four parts.

Accepted Answer

Introduction / Context: Let four AP terms be a−3d, a−d, a+d, a+3d. Their sum fixes a, and a product relation ties down d. This is a classic AP-parameterization technique for constrained partition problems. Given Data / Assumptions: Total sum = 124. AP terms: T1 = a−3d, T2 = a−d, T3 = a+d, T4 = a+3d. T1*T4 is 128 less than T2*T3. Concept / Approach: Use sum to get a, then convert the product condition into an equation in d via the identities (a−3d)(a+3d) = a^2 − 9d^2 and (a−d)(a+d) = a^2 − d^2. Step-by-Step Solution: Sum: (a−3d)+(a−d)+(a+d)+(a+3d) = 4a = 124 ⇒ a = 31Product condition: (a^2 − 9d^2) = (a^2 − d^2) − 128⇒ −9d^2 = −d^2 − 128 ⇒ 8d^2 = 128 ⇒ d^2 = 16 ⇒ d = ±4Terms: 31−12 = 19, 31−4 = 27, 31+4 = 35, 31+12 = 43 Verification / Alternative check: Products: 19*43 = 817; 27*35 = 945; indeed 817 = 945 − 128. Why Other Options Are Wrong: Other sets fail either the AP structure with correct center 31 or the exact 128 product difference. Common Pitfalls: Using sequential integers instead of AP form; mixing up which product is smaller. Final Answer: 19, 27, 35, 43

Question 17

Split 20 into four terms in AP with a product ratio 2:3: Divide 20 into four parts in arithmetic progression such that (1st × 4th) : (2nd × 3rd) = 2 : 3. Identify the four parts.

Accepted Answer

Introduction / Context: Again use the standard AP parameterization a−3d, a−d, a+d, a+3d with constraints on sum and a ratio of endpoint and middle products. The ratio eliminates the common factor (1/3)π type influences seen in geometric problems and focuses purely on algebraic relationships in a and d. Given Data / Assumptions: Total sum = 20 ⇒ 4a = 20 ⇒ a = 5. AP terms as above. (T1*T4) : (T2*T3) = 2 : 3. Concept / Approach: Use identities: T1*T4 = a^2 − 9d^2 and T2*T3 = a^2 − d^2. Form the ratio and solve for d. Step-by-Step Solution: (a^2 − 9d^2) / (a^2 − d^2) = 2/3With a = 5: (25 − 9d^2) / (25 − d^2) = 2/33(25 − 9d^2) = 2(25 − d^2) ⇒ 75 − 27d^2 = 50 − 2d^225 = 25d^2 ⇒ d^2 = 1 ⇒ d = ±1Terms: 5−3 = 2, 5−1 = 4, 5+1 = 6, 5+3 = 8 Verification / Alternative check: Products: 2*8 = 16 and 4*6 = 24; the ratio is 16:24 = 2:3 as required. Why Other Options Are Wrong: They either do not sum to 20 or are not an AP centered at 5 with common difference 1. Common Pitfalls: Trying to brute-force options without noting the symmetric AP form around the average (sum/4). Final Answer: 2, 4, 6, 8

Question 18

Savings in an arithmetic progression over 10 years: A man saves ₹ 1,45,000 in ten years. Each year after the first, he saves ₹ 2,000 more than in the previous year. How much did he save in the first year?

Accepted Answer

Introduction / Context: This is a direct application of the sum formula for an arithmetic progression (AP) where yearly savings increase by a fixed amount. Identify the first term a and common difference d, then equate the AP sum to the total savings. Given Data / Assumptions: Total years n = 10. Common difference d = ₹ 2000 per year. Total sum S = ₹ 1,45,000. Concept / Approach: For an AP, S = n/2 * [2a + (n − 1)d]. Solve for a given S, n, and d. Step-by-Step Solution: S = 10/2 * [2a + 9*2000] = 5 * (2a + 18000)145000 = 5 * (2a + 18000) ⇒ 2a + 18000 = 290002a = 11000 ⇒ a = 5500 Verification / Alternative check: List a few terms: 5500, 7500, 9500, ... (increasing by 2000) and confirm that the AP sum equals ₹ 1,45,000. Why Other Options Are Wrong: They do not satisfy the AP sum equation with n = 10 and d = ₹ 2000. Common Pitfalls: Using n rather than (n − 1) with d; forgetting to halve the product when applying S = n/2 [...]. Final Answer: ₹ 5500

Question 19

Common terms count in two APs: How many terms are common to the two sequences: AP1 = 1, 3, 5, ... (120 terms) and AP2 = 3, 6, 9, ... (80 terms)?

Accepted Answer

Introduction / Context: To count common terms of two APs, characterize the set of numbers belonging to both and count how many such numbers fall within the overlapping numeric range. Here, one AP is odd numbers; the other is multiples of 3. Given Data / Assumptions: AP1: a1 = 1, d1 = 2, n1 = 120 ⇒ last term L1 = 1 + 119*2 = 239. AP2: a2 = 3, d2 = 3, n2 = 80 ⇒ last term L2 = 3 + 79*3 = 240. Common terms are odd multiples of 3, i.e., numbers congruent to 3 mod 6. Concept / Approach: Common sequence C: 3, 9, 15, ... with common difference 6. Count how many are ≤ min(L1, L2) = 239. Step-by-Step Solution: C_k = 3 + (k−1)*63 + (k−1)*6 ≤ 239 ⇒ (k−1)*6 ≤ 236 ⇒ k−1 ≤ 39 ⇒ k ≤ 40Hence, there are 40 common terms. Verification / Alternative check: The 40th common term is 3 + 39*6 = 237, which is ≤ 239 and also in AP2 (≤ 240). Why Other Options Are Wrong: 39 undercounts (ignores the 237 term); 41+ would exceed 239. Common Pitfalls: Using 240 as the cap (AP1 ends at 239), or taking difference 3 instead of 6 for the intersection. Final Answer: 40

Question 20

Summation of product-pattern series: Find S(n) = 1·2·4 + 2·3·5 + 3·4·6 + ... up to n terms.

Accepted Answer

Introduction / Context: Each term of the series is of the form k(k+1)(k+3) for k = 1..n. Converting products into polynomial form allows use of standard summations of k, k^2, and k^3. Given Data / Assumptions: T_k = k(k+1)(k+3) for k ≥ 1. We want S(n) = Σ T_k for k = 1..n. Concept / Approach: Expand T_k = k(k^2 + 4k + 3) = k^3 + 4k^2 + 3k. Then use standard identities: Σk = n(n+1)/2, Σk^2 = n(n+1)(2n+1)/6, Σk^3 = [n(n+1)/2]^2. Step-by-Step Solution: S = Σ(k^3 + 4k^2 + 3k) = Σk^3 + 4Σk^2 + 3Σk= [n^2(n+1)^2]/4 + 4[n(n+1)(2n+1)/6] + 3[n(n+1)/2]= n(n+1)/12 * (3n^2 + 19n + 26) Verification / Alternative check: Test small n (e.g., n = 1 or 2) to confirm that both sides match numerically. Why Other Options Are Wrong: They miss coefficients from the polynomial expansion or apply an incorrect factor outside the bracket. Common Pitfalls: Algebraic slips in expanding k(k+1)(k+3) or forgetting the factor 4 in front of Σk^2. Final Answer: [n(n + 1) / 12] (3n2 + 19n + 26)

Odd Man Out and Series Questions