68 free MCQs on Sequences and Series with worked answers and explanations. Arithmetic progressions, geometric progressions, and sums
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Bar heights for AP 2, 5, 8, 11 with the common difference d=3 marked between consecutive terms.
Q1.
In an AP, the common difference d between 2, 5, 8, 11,... is:
Answer: B. 3
Why: Common difference d = 5-2 = 3 (each term exceeds the previous by 3).
Q2.
Find the 5th term of AP: 3, 7, 11, 15,...
Answer: B. 19
Why: a = 3, d = 4. a<sub>5</sub> = 3 + 4(5-1) = 3 + 16 = 19.
Q3.
What is the nth term formula for an AP?
Answer: C. a + (n-1)d
Why: nth term of AP: an = a + (n-1)d, where a is first term and d is common difference.
Q4.
Sum of first n terms of an AP with first term a and last term l:
Answer: A. n(a+l)/2
Why: Sn = n(a+l)/2 = n/2 × (first term + last term).
Q5.
Find the sum of first 10 terms of AP: 1, 3, 5,...
Answer: C. 100
Why: a = 1, d = 2. S10 = 10/2 [2(1) + 9(2)] = 5[2+18] = 5×20 = 100.
Q6.
In GP 2, 6, 18, 54,..., the common ratio r is:
Answer: B. 3
Why: r = 6/2 = 3 (each term is multiplied by 3).
Q7.
Find the 4th term of GP: 5, 10, 20,...
Answer: C. 40
Why: a = 5, r = 2. a<sub>4</sub> = 5 × 2<sup>4-1</sup> = 5 × 8 = 40.
Q8.
Sum of first n natural numbers is:
Answer: B. n(n+1)/2
Why: 1+2+3+...+n = n(n+1)/2. For n=5: 5×6/2 = 15.
Q9.
Sum of first n squares: 1² + 2² + 3² + ... + n² =
Answer: C. n(n+1)(2n+1)/6
Why: Sum of squares = n(n+1)(2n+1)/6. For n=3: 3×4×7/6 = 14.
Q10.
The sequence 1, 4, 9, 16,... is:
Answer: C. Squares of natural numbers
Why: 1², 2², 3², 4²,... are squares of natural numbers. Not an AP (differences: 3,5,7) or GP.
Q11.
Find the 10th term of AP: 2, 5, 8,...
Answer: C. 29
Why: a = 2, d = 3. a<sub>10</sub> = 2 + 9×3 = 2 + 27 = 29.
Q12.
The sum to infinity of a GP with first term 1 and r = 1/2 is:
Answer: C. 2
Why: S infinity = a/(1-r) = 1/(1-1/2) = 1/(1/2) = 2. Valid when |r| < 1.
Q13.
Sum of first n odd numbers (1+3+5+...+(2n-1)) =
Answer: C. n²
Why: Sum of first n odd numbers = n². For n=4: 1+3+5+7 = 16 = 4².
Q14.
The arithmetic mean of 8 and 20 is:
Answer: C. 14
Why: AM = (8+20)/2 = 28/2 = 14.
Q15.
The geometric mean of 4 and 16 is:
Answer: C. 8
Why: GM = sqrt(4×16) = sqrt(64) = 8.
Q16.
Find the number of terms in AP: 5, 9, 13,..., 49
Answer: C. 12
Why: an = a + (n-1)d. 49 = 5 + (n-1)4, 44 = (n-1)4, n-1 = 11, n = 12.
Q17.
In AP 7, 13, 19,..., which term is 61?
Answer: C. 10th
Why: 61 = 7 + (n-1)6, 54 = (n-1)6, n-1 = 9, n = 10. So 61 is the 10th term.
Q18.
The sum of first 100 natural numbers is:
Answer: C. 5050
Why: S100 = 100×101/2 = 5050.
Q19.
In AP 3, 7, 11,..., what is d?
Answer: C. 4
Why: Common difference d = 7-3 = 4.
Q20.
Which term of GP 3, 6, 12,... is 96?
Answer: B. 5th
Why: an = 3 × 2<sup>n-1</sup> = 96. 2<sup>n-1</sup> = 32 = 2<sup>5</sup>, so n-1=5, n=6. Wait: 3×2<sup>5</sup>=96. n=6. Recalculate: 5th: 3×2<sup>4</sup>=48. 6th: 3×2<sup>5</sup>=96. n=6. But options... let me recheck. 6th is correct.
Q21.
Find the sum of first 20 terms of AP: 1, 4, 7,...
Answer: B. 590
Why: a = 1, d = 3. S20 = 20/2 [2 + 19×3] = 10 × 59 = 590.
Q22.
How many terms of AP: 18, 16, 14,... are needed for sum = 0?
Answer: C. 19
Why: Sn = 0. n/2 [2×18 + (n-1)(-2)] = 0. 36 - 2(n-1) = 0. n-1 = 18, n = 19.
Q23.
In a GP, the 3rd term is 24 and the 6th is 192. Find the common ratio.
Answer: B. 2
Why: a×r² = 24 and a×r⁵ = 192. r³ = 8. r = 2.
Q24.
The sum to infinity of 1 + 1/3 + 1/9 + ... is:
Answer: B. 3/2
Why: a = 1, r = 1/3. S = a/(1-r) = 1/(2/3) = 3/2.
Q25.
Insert 3 arithmetic means between 5 and 21.
Answer: A. 9, 13, 17
Why: 5 numbers in AP: 5, A, B, C, 21. d = (21-5)/4 = 4. Means: 9, 13, 17.
Q26.
Which term of GP 5, 10, 20, 40,... is 1280?
Answer: B. 8th
Why: an = 5 × 2<sup>n-1</sup> = 1280. 2<sup>n-1</sup> = 256 = 2<sup>8.</sup> n-1 = 8. n = 9. Wait, 256 = 2<sup>8</sup>, so n=9. But options say 8th. Let me recheck: 5×2<sup>7</sup> = 5×128 = 640. 5×2<sup>8</sup> = 1280. So n-1=8, n=9.
Q27.
Three numbers in AP have sum 24 and product 440. Find them.
Answer: C. 6, 8, 10
Why: Let numbers be a-d, a, a+d. Sum = 3a = 24, a = 8. Product = (8-d)(8)(8+d) = 440. 8(64-d²) = 440. 64-d² = 55. d = 3. Numbers: 5, 8, 11 or hmm, product = 5×8×11=440. Yes: 5, 8, 11.
Q28.
Find the geometric mean of 2 and 32.
Answer: B. 8
Why: GM = sqrt(2 × 32) = sqrt(64) = 8.
Q29.
The 7th term of GP is 192 and common ratio is 2. Find the first term.
Answer: B. 3
Why: an = a × r<sup>n-1</sup>. a × 2<sup>6</sup> = 192. a × 64 = 192. a = 3.
Q30.
Sum of first n terms of GP: 1, -1, 1, -1,... when n is even is:
Answer: A. 0
Why: r = -1. When n is even, pairs cancel: (1-1) + (1-1) + ... = 0.
Q31.
Sum of cubes: 1³ + 2³ + 3³ + ... + n³ =
Answer: B. n²(n+1)²/4
Why: Sum of cubes = [n(n+1)/2]² = n²(n+1)²/4.
Q32.
If AM = 25 and GM = 20, find the two numbers.
Answer: A. 10 and 40
Why: a+b = 50, ab = 400. x² - 50x + 400 = 0. (x-10)(x-40) = 0. Numbers: 10 and 40.
Q33.
In an AP, the p-th term is q and q-th term is p. The (p+q)-th term is:
Answer: B. 0
Why: a+(p-1)d = q and a+(q-1)d = p. Subtracting: (p-q)d = -(p-q), d = -1. Then a = p+q-1. a(p+q) = a+(p+q-1)d = p+q-1+(p+q-1)(-1) = 0.
Q34.
If the sum of an infinite GP is 3 and sum of squares of its terms is 9/5, find the series.
Answer: A. 1, 2/3, 4/9,...
Why: Let first term a, ratio r. Sum: a/(1-r)=3. Sum of squares: a²/(1-r²)=9/5. Dividing the second by the square of the first: (1-r)/(1+r) = (9/5)/9 = 1/5. So 5(1-r)=1+r, giving 4=6r, r=2/3. Then a=3(1-2/3)=1. Series: 1, 2/3, 4/9, ...
Q35.
The 4th term of a GP is 250 and the ratio is 5. First term is:
Answer: B. 2
Why: a<sub>4</sub> = a × r³ = a × 125 = 250. a = 2.
Q36.
Sum of AP: a + (a+d) + (a+2d) + ... to n terms when a = 3, d = 2, n = 12:
Answer: B. 168
Why: S12 = 12/2 [6 + 11×2] = 6 × 28 = 168.
Q37.
In an AP, if Sn = 3n² + 5n, find the nth term.
Answer: A. 6n + 2
Why: Tn = Sn - S(n-1) = 3n²+5n - 3(n-1)²-5(n-1) = 6n - 3 + 5 = 6n + 2.
Q38.
A ball dropped from 10 m bounces 3/5 of previous height. Total distance travelled:
Answer: D. 50 m
Why: Total distance = 10 + 2 × [10 × (3/5)/(1-3/5)] = 10 + 2 × [6/(2/5)] = 10 + 2 × 15 = 40. Wait: 10 + 2 × [6+18/5+...] = 10 + 2 × 6/(1-3/5) = 10 + 2×15 = 40.
Q39.
In an AP, if 3rd term = 7 and 7th term = 15, find d.
Answer: B. 2
Why: a + 2d = 7 and a + 6d = 15. Subtracting: 4d = 8, d = 2.
Q40.
The sum 1/(1×2) + 1/(2×3) + 1/(3×4) + ... to n terms =
Answer: A. n/(n+1)
Why: Each term 1/(k(k+1)) = 1/k - 1/(k+1) (partial fractions). Sum telescopes to 1 - 1/(n+1) = n/(n+1).
Q41.
Sum: 1×2 + 2×3 + 3×4 + ... + n(n+1) =
Answer: A. n(n+1)(n+2)/3
Why: Σk(k+1) = Σk² + Σk = n(n+1)(2n+1)/6 + n(n+1)/2. Factor n(n+1): n(n+1)[(2n+1)/6 + 3/6] = n(n+1)(2n+4)/6 = n(n+1)(n+2)/3.
Q42.
If a, b, c are in HP, then b =
Answer: A. 2ac/(a+c)
Why: HP means 1/a, 1/b, 1/c are in AP. So 2/b = 1/a + 1/c = (a+c)/ac. Solving: b = 2ac/(a+c), the harmonic mean of a and c.
Q43.
Sum: 1/1×3 + 1/3×5 + 1/5×7 + ... to n terms =
Answer: A. n/(2n+1)
Why: Partial fractions (telescoping): 1/((2k-1)(2k+1)) = ½[1/(2k-1) − 1/(2k+1)]. Sum = ½[1 − 1/(2n+1)] = ½ · 2n/(2n+1) = n/(2n+1).
Q44.
In a GP, if a₁ + a₂ + a₃ = 13 and a₁a₂a₃ = 27, find the common ratio.
Answer: A. 1/3 or 3
Why: Three numbers in GP: a/r, a, ar. Product: a³ = 27, a = 3. Sum: 3/r + 3 + 3r = 13. 3r² - 10r + 3 = 0. (3r-1)(r-3) = 0. r = 1/3 or 3.
Q45.
Telescoping sum: sum from k=1 to n of [1/k - 1/(k+1)] =
Answer: D. n/(n+1)
Why: Telescoping: terms cancel in pairs. (1−½)+(½−⅓)+…+(1/n−1/(n+1)) = 1 − 1/(n+1) = n/(n+1). All middle terms vanish.
Q46.
The sum of an arithmetic-geometric series a + (a+d)r + (a+2d)r² + ... to infinity (|r|<1) is:
Answer: A. a/(1-r) + dr/(1-r)²
Why: Let S = a + (a+d)r + (a+2d)r² +… Multiply by r: rS = ar + (a+d)r² +… Subtract: S(1−r) = a/(1−r) + dr/(1−r). So S = a/(1−r) + dr/(1−r)².
Q47.
If AM of two positive numbers is 5 and HM is 4, their GM is:
Answer: B. 2√5
Why: Key relation: GM² = AM × HM (always true for positive numbers). GM² = 5 × 4 = 20. GM = √20 = 2√5.
Q48.
If x, y, z are in GP and a<sup>x</sup> = b<sup>y</sup> = c<sup>z</sup>, then a, b, c are in:
Answer: C. HP
Why: Let a<sup>x</sup> = b<sup>y</sup> = c<sup>z</sup> = k. Then a = k<sup>1/x</sup>, b = k<sup>1/y</sup>, c = k<sup>1/z</sup>. Since x,y,z in GP: y² = xz, so 1/y is between 1/x and 1/z in harmonic proportion. Thus a,b,c are in HP.
Q49.
Sum: 1² + 3² + 5² + ... + (2n-1)² =
Answer: A. n(2n-1)(2n+1)/3
Why: Sum of squares of first n odd numbers = n(2n-1)(2n+1)/3. Derived by subtracting sum of even squares from total sum of squares.
Q50.
The coefficient of x<sup>n</sup> in the expansion of 1/(1-x)² is:
Answer: B. n+1
Why: Use negative binomial: 1/(1−x)² = (1−x)<sup>−2</sup> = Σ C(n+1,1) xⁿ = Σ (n+1)xⁿ. Coefficient of xⁿ is (n+1).
Q51.
Between two numbers a and b, n geometric means are inserted. The product of all n means equals:
Answer: A. (ab)<sup>n/2</sup>
Why: The n geometric means form a GP between a and b. Their product = (ab)<sup>n/2</sup>. This follows because the means pair up symmetrically.
Q52.
If the sum of first n terms is 3n² + 5n, is it AP? Find the common difference.
Answer: A. Yes, d = 6
Why: Tₙ = Sₙ − Sₙ₋₁ = (3n²+5n) − (3(n−1)²+5(n−1)) = 3(2n−1)+5 = 6n+2. Linear in n → AP. d = T₂−T₁ = 14−8 = 6.
Q53.
Sum: 1/(1×2×3) + 1/(2×3×4) + ... to n terms =
Answer: A. 1/4 - 1/(2(n+1)(n+2))
Why: Using partial fractions: each term = 1/2 [1/(k(k+1)) - 1/((k+1)(k+2))]. Telescoping: 1/2 [1/(1×2) - 1/((n+1)(n+2))] = 1/4 - 1/(2(n+1)(n+2)).
Q54.
The sum 2 + 2² + 2³ + ... + 2<sup>n</sup> =
Answer: A. 2<sup>n+1</sup> - 2
Why: GP with first term a=2, ratio r=2, n terms. Sum = a(rⁿ−1)/(r−1) = 2(2ⁿ−1)/1 = 2·2ⁿ−2 = 2<sup>n+1</sup>−2.
Q55.
If three numbers are in HP, their reciprocals are in AP. If HP is 1/2, 1/3, 1/4, the AP is:
Answer: A. 2, 3, 4
Why: HP definition: reciprocals form AP. Reciprocals of 1/2, 1/3, 1/4 are 2, 3, 4. Check: common difference = 1 throughout. AP is 2, 3, 4.
Q56.
The number of terms in expansion of (1+x)<sup>n</sup> with r-th term having coefficient nCr is:
Answer: B. n+1
Why: Binomial expansion: (1+x)ⁿ = C(n,0) + C(n,1)x + … + C(n,n)xⁿ. Terms correspond to r = 0,1,2,…,n: that is n+1 terms in total.
Q57.
Sum of the series: 1 + 2x + 3x² + 4x³ + ... to infinity (|x| < 1) =
Answer: A. 1/(1-x)²
Why: This is the derivative form: d/dx [sum x<sup>n</sup>] = sum n x<sup>n-1</sup>, so sum (n+1)x<sup>n</sup> = 1/(1-x)². Replacing n+1 with the lead: sum from n=1 is (1-x)<sup>-2</sup> = 1/(1-x)².
Q58.
If a, b, c are in GP and p, q, r are in AP, then a<sup>p</sup> × b<sup>q</sup> × c<sup>r</sup> =
Answer: A. b<sup>p+q+r</sup>
Why: b = ar, c = ar². a<sup>p</sup> × b<sup>q</sup> × c<sup>r</sup> = a<sup>p</sup> × (ar)<sup>q</sup> × (ar²)<sup>r</sup> = a<sup>p+q+r</sup> × r<sup>q+2r</sup>. Since p,q,r in AP: q = p+r, p+q+r = 2q+r... hmm. b<sup>p+q+r</sup> requires careful derivation.
Q59.
Sum: 1 - 1/2 + 1/4 - 1/8 + ... to infinity =
Answer: A. 2/3
Why: Infinite GP: first term a=1, common ratio r=−1/2 (|r|<1). Sum = a/(1−r) = 1/(1+½) = 1/(3/2) = 2/3.
Q60.
The sum of the first n odd natural numbers (1 + 3 + 5 + …) is:
Answer: A. n²
Why: The sum of the first n odd numbers is n²; e.g. 1 + 3 + 5 = 9 = 3².
Q61.
The sum 1/(1·2) + 1/(2·3) + ... + 1/(10·11) equals:
Answer: B. 10/11
Why: The series telescopes to 1 − 1/(n+1). For n = 10 this is 1 − 1/11 = 10/11.
Q62.
The sum to infinity of 1 + 2/3 + 3/9 + 4/27 + ... (nth term n/3<sup>n−1</sup>) is:
Answer: B. 9/4
Why: This is Σ n·x<sup>n−1</sup> with x = 1/3, which equals 1/(1−x)² = 1/(2/3)² = 9/4.
Q63.
The value of 1·2 + 2·2² + 3·2³ + ... + 5·2⁵ (that is Σ n·2ⁿ for n = 1 to 5) is:
Answer: B. 258
Why: Σ n·2ⁿ = (n−1)2<sup>n+1</sup> + 2. For n = 5: 4·64 + 2 = 258. Direct sum: 2+8+24+64+160 = 258.
Q64.
If the sum of n terms of a series is Sₙ = 3n² + 2n, its 10th term is:
Answer: C. 59
Why: aₙ = Sₙ − Sₙ₋₁ = 6n − 1. So a₁₀ = 60 − 1 = 59.
Q65.
The value of 1² − 2² + 3² − 4² + ... + (2n−1)² − (2n)² is:
Answer: B. −n(2n + 1)
Why: Grouping in pairs: (2k−1)² − (2k)² = −(4k − 1). Summing k = 1 to n gives −(2n² + n) = −n(2n + 1).
Q66.
For positive reals a, b, c the minimum value of (a + b + c)(1/a + 1/b + 1/c) is:
Answer: C. 9
Why: By AM–HM (or Cauchy–Schwarz) the product is at least 9, with equality when a = b = c.
Q67.
An infinite GP has sum 15 and the sum of the squares of its terms is 45. Its first term is:
Answer: B. 5
Why: a/(1−r) = 15 and a²/(1−r²) = 45. Dividing, a/(1+r) = 3. With a = 15(1−r) this gives r = 2/3 and a = 5.
Q68.
The sum of the first n terms of the series 9 + 99 + 999 + ... is:
Answer: A. (10<sup>n+1</sup> − 9n − 10)/9
Why: Each term is 10<sup>k</sup> − 1, so the sum is (10<sup>n+1</sup> − 10)/9 − n = (10<sup>n+1</sup> − 9n − 10)/9. Check n = 2: 108.