Free Topic-Wise General Studies MCQs
Enhance your pattern recognition with Sequences and Series MCQs. High-standard logical and mathematical progression practice for the Civil Services Prelims.
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Explanation: For an AP with first term a = 25000 and common difference d = 1500, the nth term is T_n = a + (n-1)d. For n = 12, T_12 = 25000 + (12-1)×1500 = 41500. The sum of n terms is S_n = n/2 × [2a + (n-1)d]. Substituting, S_12 = 12/2 × [2×25000 + (12-1)×1500] = 399000. Since this is annual salary, total for 12 years in monthly terms is 399000 × 12 = ₹4788000.
Explanation: Use the sum formula S_n = n/2 × (a + l). Rearranging, n = 2S_n / (a + l). Substituting S_n = 10500, a = 7, l = 203, we get n = 2×10500 / (7 + 203) = 21000/210 = 100.
Explanation: Represent the three numbers in AP as (a-d), a, (a+d). Their sum is 3a = 27, so a = 9. The sum of squares is (a-d)² + a² + (a+d)² = 3a² + 2d² = 293. Substituting a = 9, we get 3×81 + 2d² = 293, giving 2d² = 50, so d = ±5. The numbers are 4, 9, 14. The largest is 14.
Explanation: The mth term is T_m = a + (m-1)d and the nth term is T_n = a + (n-1)d. Given T_7/T_3 = 5/2, we have (a + 6d)/(a + 2d) = 5/2. Cross-multiplying: 2(a + 6d) = 5(a + 2d). Substituting a = 4: 2(4 + 6d) = 5(4 + 2d). Simplifying: 8 + 12d = 20 + 10d. This gives 2d = 12, so d = 6.
Explanation: Two-digit numbers divisible by 3 form an AP: 12, 15, ..., 99. Number of terms n = (99-12)/3 + 1 = 30. Sum = 30/2 × (12+99) = 1665. Numbers divisible by both 3 and 5 (i.e., by 15) form an AP: 15, 30, ..., 90. Number of terms = (90-15)/15 + 1 = 6. Sum = 6/2 × (15+90) = 315. The required sum is 1665 - 315 = 1350.
Explanation: When inserting 4 arithmetic means between 5 and 25, the sequence has 6 terms with common difference d = (last - first)/(number of gaps) = (25-5)/5 = 4. The means are 9, 13, 17, 21. The third mean is 17.
Explanation: Given S_p = q and S_q = p for an AP. Using S_n = n/2 × [2a + (n-1)d], we get two equations: p/2 × [2a + (p-1)d] = 20 and q/2 × [2a + (q-1)d] = 10. Solving these simultaneously yields the common difference d = -2(p+q)/(pq) = -0.300 and first term a = 0.650. Then S_{p+q} = (p+q)/2 × [2a + (p+q-1)d]. Substituting the values of a and d gives S_{30} = -(p+q) = -30.
Explanation: For two APs, the ratio of their mth terms equals the ratio of their (2m-1)th sums. This is because T_m = S_m - S_{m-1}, and for the ratio to simplify elegantly, we use the property that the ratio of mth terms = (2m-1)th sum ratio. For the 10th term, we substitute n = 2×10-1 = 19 into the given ratio: (3×19+5):(5×19+3) = 62:98 = 31:49.
Explanation: The nth term is T_n = a + (n-1)d = 25 + (n-1)×(-3) = 25 - 3(n-1). For the first negative term, T_n < 0, so 25 - 3(n-1) < 0, which gives 3(n-1) > 25, or n-1 > 8.33. Thus n > 9.33. Since n must be an integer, the first negative term is the 10th term. Verification: T_9 = 1 (positive), T_10 = -2 (negative).
Explanation: Let the four angles in AP be (a-3d), (a-d), (a+d), (a+3d). Their sum is 4a = 360°, so a = 90°. Given the ratio of smallest to largest is 1:5, we have (a-3d)/(a+3d) = 1/5. Cross-multiplying: 5(a-3d) = a+3d, so 5a-15d = a+3d, giving 4a = 18d, thus d = 20°. The angles are 30°, 70°, 110°, 150°. The second largest is 110°.
Explanation: This represents geometric growth. If the initial amount is a = 500, and it doubles every hour (r = 2), the amount after n hours is given by a × r^n. After 6 hours, the population is 500 × 2^6 = 500 × 64 = 32000.
Explanation: This is an infinite geometric series with first term a = 27 and common ratio r = 1/3. Since |r| < 1, the sum converges to S_∞ = a/(1-r) = 27/(1 - 1/3) = 27/(2/3) = 81/2 = 40.5.
Explanation: In a GP, T_n = ar^(n-1). Given T_3 = ar² = 24 and T_6 = ar⁵ = 192. Dividing the second equation by the first: T_6/T_3 = r³ = 192/24 = 8. Therefore, r = ∛8 = 2.
Explanation: Let the three numbers in GP be a/r, a, ar. Their product is (a/r)×a×(ar) = a³ = 729, so a = 9. Their sum is a/r + a + ar = a(1/r + 1 + r) = 39. Substituting a = 9: 9(1 + r + r²)/r = 39. This simplifies to 9r² - 39r + 9 = 0, giving r = 3 or 1/3. The numbers are 3, 9, 27. The largest is 27.
Explanation: Depreciation at 10% per year means the value each year is 90% of the previous year. This forms a GP with a = 50000 and r = 0.9. The value after n years is ar^n = 50000 × (0.9)^4 = 50000 × 0.6561 = ₹32805.
Explanation: For a GP with first term a = 3 and common ratio r = 2, the sum of n terms is S_n = a(r^n - 1)/(r - 1) for r > 1. Substituting n = 8: S_8 = 3×(2^8 - 1)/(2-1) = 3×(255)/1 = 765.
Explanation: Let the GP be a, ar, ar², so b = ar and c = ar². Since a, x, b are in AP, x = (a+b)/2 = a(1+r)/2. Since b, y, c are in AP, y = (b+c)/2 = ar(1+r)/2. Then 1/x = 2/[a(1+r)] and 1/y = 2/[ar(1+r)]. Adding: 1/x + 1/y = 2/[a(1+r)] × (1 + 1/r) = 2/[a(1+r)] × (r+1)/r = 2/(ar) = 2/b.
Explanation: The decimal 0.232323... can be written as the infinite GP: 23/100 + 23/10000 + 23/1000000 + ... with first term a = 23/100 and common ratio r = 1/100. The sum is S_∞ = a/(1-r) = (23/100)/(1 - 1/100) = (23/100)/(99/100) = 23/99. Thus p = 23 and q = 99, so p + q = 122.
Explanation: For a GP with first term a = 81 and common ratio r = -1/3, the nth term is T_n = ar^(n-1). The 5th term is T_5 = 81 × (-1/3)^4 = 81 × 0.012346 = 0.
Explanation: When inserting 3 geometric means between 1 and 16, the sequence has 5 terms. The common ratio r satisfies ar^4 = 16, so r^4 = 16/1 = 16, giving r = 2. The means are 2, 4, 8. The second geometric mean is 4.
Explanation: The sequence is an arithmetic progression with common difference d = 5. Each term increases by 5. The term before the missing position is 23, so the missing term is 23 + 5 = 28.
Explanation: The sequence is a geometric progression with common ratio r = 3. Each term is multiplied by 3. The term before the missing position is 54, so the missing term is 54 × 3 = 162.
Explanation: This is a second-order sequence. The first differences are 5, 11, 17, 23, 29, 35. The second differences are constant at 6. The first difference before the missing term should be 23 + 6 = 29. Thus the missing term is 34 + 29 = 57.
Explanation: This sequence consists of two interleaved arithmetic progressions. The odd-positioned terms (1st, 3rd, 5th, ...) are 2, 5, 8, 11, 14 with common difference 3. The even-positioned terms (2nd, 4th, 6th, ...) are 3, 7, 11, 15, ? with common difference 4. The missing term is the 10th term (even position), so it is 15 + 4 = 19.
Explanation: The sequence follows the pattern n³ + 1. For n = 1, 2, 3, 4, 5, 6, 7, the terms are 1³+1=2, 2³+1=9, 3³+1=28, 4³+1=65, 5³+1=126, 6³+1=217, 7³+1=344. The missing term is the 4th term: 4³ + 1 = 64 + 1 = 65.
Explanation: The sequence follows the pattern n² + n (or n(n+1)). For n = 1, 2, 3, 4, 5, 6, 7, the terms are 1×2=2, 2×3=6, 3×4=12, 4×5=20, 5×6=30, 6×7=42, 7×8=56. The missing term is the 5th term: 5×6 = 30.
Explanation: The sequence is generated by taking consecutive prime numbers and squaring them, then subtracting 1. The primes are 2, 3, 5, 7, 11, 13, giving 2²-1=3, 3²-1=8, 5²-1=24, 7²-1=48, 11²-1=120, 13²-1=168. The missing term corresponds to the 4th prime (7): 7² - 1 = 48.
Explanation: This is a Fibonacci-type sequence where each term (from the 3rd onward) is the sum of the two preceding terms. Starting with 1 and 3: 1+3=4, 3+4=7, 4+7=11, 7+11=18, 11+18=29, 18+29=47, 29+47=76. The missing term is 7 + 11 = 18.
Explanation: The sequence follows the pattern n! + 1 (factorial plus one). For n = 1, 2, 3, 4, 5, 6, 7: 1!+1=2, 2!+1=3, 3!+1=7, 4!+1=25, 5!+1=121, 6!+1=721, 7!+1=5041. The missing term is the 4th term: 4! + 1 = 24 + 1 = 25.
Explanation: The sequence follows the pattern 2^n - 1 (Mersenne numbers). For n = 1, 2, 3, 4, 5, 6, 7: 2¹-1=1, 2²-1=3, 2³-1=7, 2⁴-1=15, 2⁵-1=31, 2⁶-1=63, 2⁷-1=127. The missing term is the 5th term: 2⁵ - 1 = 32 - 1 = 31.
Explanation: The sequence consists of triangular numbers given by n(n+1)/2. For n = 1, 2, 3, 4, 5, 6, 7, 8: 1×2/2=1, 2×3/2=3, 3×4/2=6, 4×5/2=10, 5×6/2=15, 6×7/2=21, 7×8/2=28, 8×9/2=36. The missing term is the 6th triangular number: 6×7/2 = 21.
Explanation: This is an alternating-sign arithmetic progression. The absolute values form an AP: 50, 45, 40, 35, 30, 25, 20 with common difference 5. The signs alternate starting with positive. The missing term is the 5th term, which is positive (odd position), and its absolute value is 50 - 4×5 = 30. Thus the missing term is 30.
Explanation: The sequence is a geometric progression with common ratio r = 1/3. Each term is one-third of the previous term. The term before the missing position is 81, so the missing term is 81 ÷ 3 = 27.
Explanation: The differences between consecutive terms form the sequence 1, 3, 7, 15, 31, 63, which follows the pattern 2^n - 1. The difference before the missing term is 2⁴ - 1 = 15. Thus the missing term is 12 + 15 = 27.
Explanation: Starting from the third term, each term is the sum of the two preceding terms minus 1. Verification: 5+7-1=11, 7+11-1=17, 11+17-1=27, 17+27-1=43, 27+43-1=69. The missing term is 11 + 17 - 1 = 27.
Explanation: For two numbers a and b, AM = (a+b)/2 = 25, so a+b = 50. GM = √(ab) = 15, so ab = 225. The numbers are roots of x² - (a+b)x + ab = 0, i.e., x² - 50x + 225 = 0. Factoring: (x-5)(x-45) = 0. The numbers are 5 and 45. The smaller number is 5.
Explanation: Since a, b, c are in AP, the middle term is the arithmetic mean: 2b = a + c, so c = 2b - a = 2×4 - 2 = 6. Since a, b, d are in GP, the middle term is the geometric mean: b² = ad, so d = b²/a = 16/2 = 8. Therefore, c + d = 6 + 8 = 14.
Explanation: Each term can be written as 1/[k(k+1)] = 1/k - 1/(k+1). The sum becomes (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/15 - 1/16). This is a telescoping series where all intermediate terms cancel, leaving 1 - 1/16 = 15/16.
Explanation: This is a geometric series with first term a = 1 and common ratio r = 2/3. Since |r| < 1, the sum of n terms is S_n = a(1-r^n)/(1-r). For n = 6: S_6 = 1×(1-(2/3)^6)/(1-2/3) = (1-(2/3)^6)/(1/3) = 3×(1-(2/3)^6) = 665/243.
Explanation: The products of corresponding terms are: 2×3=6, 5×6=30, 8×12=96, 11×24=264, 14×48=672. The sum is 6 + 30 + 96 + 264 + 672 = 1068.
Explanation: Using logarithm properties, log₂(2) = 1, log₂(8) = 3, log₂(32) = 5, log₂(128) = 7, log₂(512) = 9. The expression becomes 1 + 3 + 5 + 7 + 9, which is an AP with first term 1, common difference 2, and 5 terms. The sum is 5/2 × (2×1 + 4×2) = 5/2 × 10 = 25.
Explanation: By the AM-GM inequality, for any positive real number x, the arithmetic mean of x and 1/x is greater than or equal to their geometric mean: (x + 1/x)/2 ≥ √(x × 1/x) = 1. Therefore, x + 1/x ≥ 2. The minimum value 2 is achieved when x = 1/x, i.e., x = 1.
Explanation: In any AP, the sum of terms equidistant from the beginning and end is constant and equals the sum of the first and last terms. Thus T_3 + T_15 = T_1 + T_17 = 40. The sum of n terms is S_n = n/2 × (first + last). Therefore, S_17 = 17/2 × 40 = 340.
Explanation: Using the GP sum formula S_n = a(r^n - 1)/(r - 1) for r > 1. Substituting S_5 = 242, a = 2, n = 5: 242 = 2(r^5 - 1)/(r - 1). This gives (r^5 - 1)/(r - 1) = 121. Since (r^5 - 1)/(r - 1) = 1 + r + r² + r³ + r⁴, we test r = 3: 1 + 3 + 9 + 27 + 81 = 121. Thus r = 3.
Explanation: Given T_n = ar^(n-1) = 96 and S_n = a(r^n - 1)/(r - 1) = 189. From the first equation, r^(n-1) = 96/3 = 32. From the second, (r^n - 1)/(r - 1) = 63. Since r^n = r × r^(n-1) = 2 × 32 = 64, we substitute: (64 - 1)/(r - 1) = 63. Solving gives r = 2. Then r^(n-1) = 32 gives 2^(n-1) = 32, so n-1 = 5 and n = 6.
Explanation: Let a, b, c be in GP with common ratio r, so b = ar and c = ar². Then log b = log a + log r and log c = log a + 2 log r. The differences are log b - log a = log r and log c - log b = log r, which are equal. Therefore, log a, log b, log c form an arithmetic progression with common difference log r.
Explanation: Let a, b, c be in AP with common difference d, so b = a + d and c = a + 2d. Then 2^b / 2^a = 2^(b-a) = 2^d, and 2^c / 2^b = 2^(c-b) = 2^d. Since the ratio between consecutive terms is constant (2^d), the sequence 2^a, 2^b, 2^c forms a geometric progression.
Explanation: The ball first falls 20m. Then it travels up and down: 2×15m + 2×11m + ... This is an infinite GP with first term 2rh = 30.0 and ratio r = 3/4. The sum of this GP is 2rh/(1-r) = 30.0/0.25 = 120. Adding the initial drop: total distance = 20 + 120 = 140 metres.
Explanation: Each term can be written as 5 × (1, 11, 111, 1111). Using 111...1 (k times) = (10^k - 1)/9, the sum is 5 × [(10-1)/9 + (100-1)/9 + (1000-1)/9 + (10000-1)/9] = 5/9 × [(10+100+1000+10000) - 4] = 5/9 × 11110 - 20/9 = 6170. Alternatively, direct addition: 5 + 55 + 555 + 5555 = 6170.
Explanation: Given AM = (a+b)/2 = 10, so a+b = 20. Given GM = √(ab) = 8, so ab = 64. The sum of squares is a² + b² = (a+b)² - 2ab = 20² - 2×64 = 400 - 128 = 272.
Explanation: Compound interest follows geometric progression. The amount after n years is A = P(1+r)^n = 8000×(1.1)^3 = 8000×1.331 = ₹10648. The compound interest is A - P = 10648 - 8000 = ₹2648.
Explanation: Population growth follows a geometric progression: P_n = P_0(1+r)^n. Setting 60832 = 50000 × (1.04)^n gives (1.04)^n = 1.21664. Since 1.04^5 = 1.2166529..., it takes exactly 5 years.
Explanation: The present value of an installment forms a geometric series. Cash price = x/(1+r) + x/(1+r)² + x/(1+r)³ = x/(1+r) × [1 - (1+r)^(-3)]/[1 - 1/(1+r)] = x × [1 - (1+r)^(-3)]/r. Substituting x = 4000, r = 0.1: PV = 4000 × (1 - 1.1^(-3))/0.1 = 4000 × 2.4869 = ₹9947.
Explanation: The sum of squares of the first n natural numbers is given by the formula Σn² = n(n+1)(2n+1)/6. For n = 15: Σ15² = 15×16×31/6 = 7440//6 = 1240.
Explanation: The sum of cubes of the first n natural numbers equals the square of the sum of the first n natural numbers: Σn³ = [n(n+1)/2]². For n = 10: Σ10³ = [10×11/2]² = [55]² = 3025.
Explanation: This is an arithmetico-geometric series where the coefficients form an AP (1, 2, 3, 4, 5) and the powers form a GP (2⁰, 2¹, 2², 2³, 2⁴). The sum can be computed directly: 1 + 4 + 12 + 32 + 80 = 129. Alternatively, using the formula S = [1 - (n+1)x^n + nx^(n+1)]/(1-x)² with x=2, n=5: [1 - 6×32 + 5×64]/(1-2)² = 129.
Explanation: Each term can be written as 1/[k(k+1)] = 1/k - 1/(k+1). The series becomes (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/9 - 1/10). This is a telescoping series where all intermediate terms cancel, leaving 1 - 1/10 = 9/10 = 9/10.
Explanation: This is an infinite geometric series with first term a = 1 and common ratio r = 1/3. Since |r| < 1, the sum converges to S_∞ = a/(1-r) = 1/(1 - 1/3) = 1/(2/3) = 3/2.
Explanation: Converting to fractions: 0.7 = 7/10, 0.77 = 77/100, 0.777 = 777/1000, 0.7777 = 7777/10000. With common denominator 10000: 7000/10000 + 7700/10000 + 7770/10000 + 7777/10000 = 30247/10000. Alternatively, using 0.777...7 (k times) = 7/9 × (1 - 10^(-k)), the sum is 7/9 × [4 - (0.1 + 0.01 + 0.001 + 0.0001)] = 7/9 × [4 - 0.1111] = 30247/10000.
Explanation: The general term is k(k+1) = k² + k. The sum Σk(k+1) = Σk² + Σk = n(n+1)(2n+1)/6 + n(n+1)/2 = n(n+1)/2 × [(2n+1)/3 + 1] = n(n+1)(n+2)/3. For n = 8: 8×9×10/3 = 720//3 = 240.