JEE Mathematics Sequence and Series MCQs Set 32

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Multiple Choice Questions

Question 1: In an AP with first term a, the sum of the first p terms is zero. The sum of the next q terms is:
(a) \( -\frac{aq(p-q)}{p+1} \)
(b) \( -\frac{aq(p+q)}{p-1} \)
(c) \( \frac{aq(p+q)}{p-1} \)
(d) \( -\frac{ap(p+q)}{q-1} \)
Show Answer & Explanation

Answer: (b) \( -\frac{aq(p+q)}{p-1} \)

Explanation:
1. \( S_p = 0 \Rightarrow 2a + (p-1)d = 0 \Rightarrow d = -\frac{2a}{p-1} \).
2. Sum of the next q terms: \( S_{p+q} - S_p = S_{p+q} = \frac{p+q}{2}[2a + (p+q-1)d] = \frac{p+q}{2} \cdot qd = -\frac{aq(p+q)}{p-1} \).

Teacher's Note:
1. 2a + (p-1)d = 0 simplifies the second bracket.
2. Use p = 3, a = 2 (d = -2): sum of next 1 term = a_4 = -4; formula gives -2 x 1 x 4/2 = -4.

Question 2: Divide 20 into four parts in AP such that the product of the first and the fourth is to the product of the second and the third as 2 : 3. The parts are:
(a) \( 2, 4, 6, 8 \)
(b) \( 1, 3, 7, 9 \)
(c) \( 3, 4, 6, 7 \)
(d) \( 0, 3, 7, 10 \)
Show Answer & Explanation

Answer: (a) \( 2, 4, 6, 8 \)

Explanation:
1. Let the parts be \( a - 3d, a - d, a + d, a + 3d \); \( 4a = 20 \Rightarrow a = 5 \).
2. \( \frac{25 - 9d^2}{25 - d^2} = \frac23 \Rightarrow 75 - 27d^2 = 50 - 2d^2 \Rightarrow d^2 = 1 \).
3. The parts are 2, 4, 6, 8 (16 : 24 = 2 : 3).

Teacher's Note:
1. Use the symmetric form for four terms.
2. Take d positive for increasing order.

Question 3: The sum of n terms of the series \( 1 + 4 + 10 + 22 + \cdots \) is:
(a) \( 3 \cdot 2^n - 2n - 2 \)
(b) \( 2^{n+1} - n - 1 \)
(c) \( 3 \cdot 2^{n} - n - 3 \)
(d) \( 3 \cdot 2^n - 2n - 3 \)
Show Answer & Explanation

Answer: (d) \( 3 \cdot 2^n - 2n - 3 \)

Explanation:
1. The successive differences 3, 6, 12, ... form a GP with ratio 2, so \( T_n = 3 \cdot 2^{n-1} - 2 \) (check: 1, 4, 10, 22).
2. \( S_n = 3(2^n - 1) - 2n = 3 \cdot 2^n - 2n - 3 \).

Teacher's Note:
1. Use the method of differences when the differences form an AP or GP.
2. Test with n = 2: 3 x 4 - 4 - 3 = 5.

Question 4: Insert 3 geometric means between \( 2\tfrac14 \) and \( \tfrac49 \). The means are:
(a) \( 2,\ \frac32,\ 1 \)
(b) \( \frac32,\ \frac{5}{4},\ \frac23 \)
(c) \( \frac32,\ 1,\ \frac23 \)
(d) \( \frac43,\ 1,\ \frac12 \)
Show Answer & Explanation

Answer: (c) \( \frac32,\ 1,\ \frac23 \)

Explanation:
1. \( r^4 = \frac{4/9}{9/4} = \frac{16}{81} \Rightarrow r = \frac23 \).
2. Means: \( \frac94 \cdot \frac23 = \frac32,\ \frac32 \cdot \frac23 = 1,\ 1 \cdot \frac23 = \frac23 \).

Teacher's Note:
1. Three means means five terms, so the ratio is the fourth root.
2. The middle mean is sqrt(9/4 x 4/9) = 1.

Question 5: The sum of all three-digit natural numbers divisible by 7 is:
(a) \( 69972 \)
(b) \( 70336 \)
(c) \( 70300 \)
(d) \( 70686 \)
Show Answer & Explanation

Answer: (b) \( 70336 \)

Explanation:
1. The first multiple is 105 and the last is 994, so \( n = \frac{994 - 105}{7} + 1 = 128 \).
2. Sum = \( \frac{128}{2}(105 + 994) = 64 \times 1099 = 70336 \).

Teacher's Note:
1. Find the first and last multiples inside the range.
2. Use S = n/2 (a + l).

Question 6: Find the sum of \( -2,\ 2\tfrac12,\ -3\tfrac18, \ldots \) to 6 terms.
(a) \( \frac{1281}{512} \)
(b) \( - \frac{1281}{512} \)
(c) \( - \frac{6405}{2048} \)
(d) \( \frac{1793}{512} \)
Show Answer & Explanation

Answer: (a) \( \frac{1281}{512} \)

Explanation:
1. \( a = -2,\ r = -\frac54 \) (since \( 2.5 / (-2) = -1.25 \)).
2. \( S_6 = \frac{-2[(-\frac54)^6 - 1]}{-\frac54 - 1} = \frac{1281}{512} \).

Teacher's Note:
1. The ratio is negative: the terms alternate in sign.
2. Compute (5/4)^6 = 15625/4096.

Question 7: If \( a_1, a_2, a_3, \ldots \) is an AP and \( a_1 + a_5 + a_{10} + a_{15} + a_{20} + a_{24} = 225 \), then \( a_1 + a_2 + \cdots + a_{24} \) equals:
(a) \( 675 \)
(b) \( 450 \)
(c) \( 1125 \)
(d) \( 900 \)
Show Answer & Explanation

Answer: (d) \( 900 \)

Explanation:
1. The six terms pair up as \( (a_1 + a_{24}), (a_5 + a_{20}), (a_{10} + a_{15}) \), each pair summing to the same value \( a_1 + a_{24} \).
2. So \( 3(a_1 + a_{24}) = 225 \), \( a_1 + a_{24} = 75 \).
3. \( S_{24} = 12 (a_1 + a_{24}) = 900 \).

Teacher's Note:
1. Equidistant terms of an AP have a constant sum.
2. Sum of 24 terms = 12 pairs.

Question 8: In a GP the first term is 7, the last term is 448 and the sum is 889. The common ratio is:
(a) \( 4 \)
(b) \( \frac12 \)
(c) \( 2 \)
(d) \( 3 \)
Show Answer & Explanation

Answer: (c) \( 2 \)

Explanation:
1. \( \frac{448r - 7}{r - 1} = 889 \Rightarrow 441r = 882 \Rightarrow r = 2 \).

Teacher's Note:
1. Eliminate n using the last term.
2. 7, 14, 28, 56, 112, 224, 448: sum 889.

Question 9: The rth mean between x and 2y is the same as the rth mean between 2x and y, n means being inserted in each case. The relation between x and y is:
(a) \( x = y \)
(b) \( x : y = r : (n + 1 - r) \)
(c) \( x : y = n : r \)
(d) \( x : y = (n + 1 - r) : r \)
Show Answer & Explanation

Answer: (b) \( x : y = r : (n + 1 - r) \)

Explanation:
1. The rth mean between x and 2y is \( x + \frac{r(2y - x)}{n+1} \); between 2x and y it is \( 2x + \frac{r(y - 2x)}{n+1} \).
2. Equating: \( \frac{r(2y - x) - r(y - 2x)}{n+1} = x \Rightarrow r(x + y) = (n+1)x \).
3. So \( ry = (n + 1 - r)x \), that is \( x : y = r : (n + 1 - r) \).

Teacher's Note:
1. Write the rth mean as first term + r d.
2. Check with n = 1, r = 1: x = y.

Question 10: The nth term of the sequence \( 1, 5, 13, 29, 61, \ldots \) is:
(a) \( 2^{n+1} - 3 \)
(b) \( 2^n - 1 \)
(c) \( 2^{n+1} - 2n \)
(d) \( 3 \cdot 2^{n-1} - 2 \)
Show Answer & Explanation

Answer: (a) \( 2^{n+1} - 3 \)

Explanation:
1. The differences 4, 8, 16, 32 form a GP with ratio 2, so the nth term has the form \( A \cdot 2^n + B \).
2. Using \( T_1 = 1 \) and \( T_2 = 5 \): \( 2A + B = 1,\ 4A + B = 5 \Rightarrow A = 2,\ B = -3 \). So \( T_n = 2^{n+1} - 3 \).

Teacher's Note:
1. Check n = 3: 16 - 3 = 13, and n = 5: 64 - 3 = 61.
2. The sum of n terms is 2^(n+2) - 3n - 4.

Question 11: The third term of an AP is 18 and the seventh term is 30. The sum of 17 terms is:
(a) \( 594 \)
(b) \( 630 \)
(c) \( 510 \)
(d) \( 612 \)
Show Answer & Explanation

Answer: (d) \( 612 \)

Explanation:
1. \( 4d = 12 \Rightarrow d = 3,\ a = 12 \).
2. The 9th term (middle of 17 terms) is \( 12 + 8 \cdot 3 = 36 \), and \( S_{17} = 17 \times 36 = 612 \).

Teacher's Note:
1. Sum of odd number of terms = n x middle term.
2. Alternatively use S = 17/2 (24 + 48).

Question 12: One of the roots of \( 2000x^6 + 100x^5 + 10x^3 + x - 2 = 0 \) is of the form \( \dfrac{m + \sqrt n}{r} \), where m is a non-zero integer and n, r are relatively prime natural numbers. The value of \( m + n + r \) is:
(a) \( 161 + 40 \)
(b) \( 120 \)
(c) \( 200 \)
(d) \( 161 \)
Show Answer & Explanation

Answer: (c) \( 200 \)

Explanation:
1. The polynomial factors as \( (20x^2 + x - 2)(100x^4 + 10x^2 + 1) \); the quartic factor has no real roots.
2. \( 20x^2 + x - 2 = 0 \Rightarrow x = \frac{-1 \pm \sqrt{161}}{40} \).
3. So \( m = -1,\ n = 161,\ r = 40 \) and \( m + n + r = 200 \).

Teacher's Note:
1. Group the terms: 2000x^6 + 100x^5 + 10x^3 + x - 2 factorises neatly.
2. 161 = 7 x 23 has no common factor with 40.

Question 13: Insert 6 geometric means between 14 and \( -\tfrac{7}{64} \). The common ratio of the resulting GP is:
(a) \( -2 \)
(b) \( -\frac12 \)
(c) \( \frac12 \)
(d) \( -\frac14 \)
Show Answer & Explanation

Answer: (b) \( -\frac12 \)

Explanation:
1. There are 8 terms, so \( r^7 = \frac{-7/64}{14} = -\frac{1}{128} \).
2. \( r = -\frac12 \) (odd root of a negative number).

Teacher's Note:
1. Seven steps from first to last term.
2. The means are -7, 7/2, -7/4, 7/8, -7/16, 7/32.

Question 14: The first term of an AP is 1 and the sum of the first nine terms is 369. The first and ninth terms of a GP coincide with those of the AP. The seventh term of the GP is:
(a) \( 27 \)
(b) \( 81 \)
(c) \( 9 \)
(d) \( \sqrt3 \cdot 9 \)
Show Answer & Explanation

Answer: (a) \( 27 \)

Explanation:
1. \( S_9 = \frac92(2 + 8d) = 369 \Rightarrow d = 10 \), so the ninth term is 81.
2. GP: \( ar^8 = 81 \) with \( a = 1 \), that is \( r^8 = 81 \) and \( r^2 = 3 \).
3. Seventh term: \( r^6 = (r^2)^3 = 27 \).

Teacher's Note:
1. r^8 = 81 gives r^2 = 3 (taking the positive value).
2. Only even powers of r are needed, so the sign of r does not matter.

Question 15: Find the sum of \( 2, -4, 8, \ldots \) to 10 terms.
(a) \( 682 \)
(b) \( -1022 \)
(c) \( 1022 \)
(d) \( -682.0 \)
Show Answer & Explanation

Answer: (d) \( -682.0 \)

Explanation:
1. \( a = 2,\ r = -2 \).
2. \( S_{10} = \frac{2[(-2)^{10} - 1]}{-2 - 1} = \frac{2 \times 1023}{-3} = -682 \).

Teacher's Note:
1. An even power of -2 is positive.
2. The sum is negative because the last term is -1024.

Question 16: If the third and fourth terms of an arithmetic sequence are increased by 3 and 8 respectively, the first four terms form a geometric sequence (the sequence is non-constant and has no zero term after the change). The sum of the first four terms of the original AP is:
(a) \( 60 \)
(b) \( 36 \)
(c) \( 54 \)
(d) \( 45 \)
Show Answer & Explanation

Answer: (c) \( 54 \)

Explanation:
1. Let the AP be \( a, a+d, a+2d, a+3d \). The GP condition on \( a, a+d, a+2d+3, a+3d+8 \) gives \( (a+d)^2 = a(a+2d+3) \) and \( (a+2d+3)^2 = (a+d)(a+3d+8) \).
2. Solving, \( a = 27,\ d = -9 \) (the solution \( a = 3, d = -3 \) makes the GP contain zeros). The GP is 27, 18, 12, 8 with ratio 2/3.
3. Original AP: 27, 18, 9, 0 with sum 54.

Teacher's Note:
1. Check each algebraic solution for validity.
2. The second term of the GP is 18.

Question 17: The condition that the roots of \( x^3 - px^2 + qx - r = 0 \) are in AP is \( 2p^3 - 9pq + 27r = 0 \). The roots of \( x^3 - 12x^2 + 39x - 28 = 0 \) are:
(a) \( 4, 4, 4 \)
(b) \( 1, 4, 7 \)
(c) \( 2, 4, 6 \)
(d) \( 1, 3, 8 \)
Show Answer & Explanation

Answer: (b) \( 1, 4, 7 \)

Explanation:
1. Here \( p = 12,\ q = 39,\ r = 28 \): \( 2 \cdot 1728 - 9 \cdot 12 \cdot 39 + 27 \cdot 28 = 3456 - 4212 + 756 = 0 \), so the roots are in AP.
2. The middle root is \( \frac p3 = 4 \). Then \( (4 - d)(4 + d) \cdot 4 = 28 \Rightarrow d^2 = 9 \), so the roots are 1, 4, 7.

Teacher's Note:
1. For roots in AP the middle root is p/3.
2. Check: product 28 and pair sum 4 + 7 + 28 = 39.

Question 18: Find the sum to infinity of \( 7, \sqrt{42}, 6, \ldots \)
(a) \( 49 + 7\sqrt{42} \)
(b) \( 49 - 7\sqrt{42} \)
(c) \( 7 + \sqrt{42} \)
(d) \( \frac{7}{\sqrt7 - \sqrt6} \)
Show Answer & Explanation

Answer: (a) \( 49 + 7\sqrt{42} \)

Explanation:
1. \( r = \frac{\sqrt{42}}{7} = \sqrt{\frac67} \).
2. \( S_\infty = \frac{7}{1 - \sqrt{6/7}} = \frac{7\sqrt7}{\sqrt7 - \sqrt6} = 7\sqrt7(\sqrt7 + \sqrt6) = 49 + 7\sqrt{42} \).

Teacher's Note:
1. Since 7 - 6 = 1, rationalising is simple.
2. Check: about 49 + 45.37 = 94.4.

Question 19: The sum to infinity of \( 1 + \tfrac32 + \tfrac54 + \tfrac78 + \cdots \) is:
(a) \( 4 \)
(b) \( 5 \)
(c) \( 3 \)
(d) \( 6 \)
Show Answer & Explanation

Answer: (d) \( 6 \)

Explanation:
1. Let \( S = 1 + \frac32 + \frac54 + \frac78 + \cdots \); then \( \frac S2 = \frac12 + \frac34 + \frac58 + \cdots \).
2. \( S - \frac S2 = 1 + 1 + \frac12 + \frac14 + \cdots = 1 + 2 = 3 \).
3. Hence \( S = 6 \).

Teacher's Note:
1. Subtract the series multiplied by the common ratio.
2. The nth term is (2n - 1)/2^(n-1).

Question 20: Find the sum of \( \tfrac12, \tfrac13, \tfrac29, \ldots \) to 7 terms.
(a) \( \frac{2059}{2187} \)
(b) \( \frac{7}{2} \)
(c) \( \frac{2059}{1458} \)
(d) \( \frac{2059}{729} \)
Show Answer & Explanation

Answer: (c) \( \frac{2059}{1458} \)

Explanation:
1. \( a = \frac12,\ r = \frac23 \).
2. \( S_7 = \frac{\frac12[1 - (\frac23)^7]}{\frac13} = \frac32 \times \frac{2059}{2187} = \frac{2059}{1458} \).

Teacher's Note:
1. Use S = a(1 - r^n)/(1 - r) when |r| < 1.
2. 2^7 = 128 and 3^7 = 2187.

Question 21: Find the sum of \( 1.3,\ -3.1,\ -7.5, \ldots \) to 10 terms.
(a) \( -229 \)
(b) \( -185 \)
(c) \( -145 \)
(d) \( 185 \)
Show Answer & Explanation

Answer: (b) \( -185 \)

Explanation:
1. \( a = 1.3,\ d = -4.4 \).
2. \( S_{10} = 5[2.6 + 9(-4.4)] = 5 \times (-37) = -185 \).

Teacher's Note:
1. d = -3.1 - 1.3 = -4.4.
2. Use the standard sum formula.

Question 22: Find the sum of \( \tfrac34, 1\tfrac12, 3, \ldots \) to 8 terms.
(a) \( \frac{765}{4} \)
(b) \( \frac{765}{8} \)
(c) \( \frac{259}{4} \)
(d) \( 192 \)
Show Answer & Explanation

Answer: (a) \( \frac{765}{4} \)

Explanation:
1. \( a = \frac34,\ r = 2 \).
2. \( S_8 = \frac34(2^8 - 1) = \frac{765}{4} \).

Teacher's Note:
1. Ratio 2: S = a(r^n - 1)/(r - 1).
2. 255 x 3/4 = 191.25.

Question 23: The sum to n terms of \( \dfrac{1}{1 \cdot 4 \cdot 7} + \dfrac{1}{4 \cdot 7 \cdot 10} + \dfrac{1}{7 \cdot 10 \cdot 13} + \cdots \) is:
(a) \( \frac{n}{(3n+1)(3n+4)} \)
(b) \( \frac{n(3n+5)}{4(3n+1)(3n+4)} \)
(c) \( \frac{3n+5}{8(3n+1)(3n+4)} \)
(d) \( \frac{n(3n+5)}{8(3n+1)(3n+4)} \)
Show Answer & Explanation

Answer: (d) \( \frac{n(3n+5)}{8(3n+1)(3n+4)} \)

Explanation:
1. The nth term is \( \frac{1}{(3n-2)(3n+1)(3n+4)} = \frac16\left[\frac{1}{(3n-2)(3n+1)} - \frac{1}{(3n+1)(3n+4)}\right] \).
2. The sum telescopes: \( S_n = \frac16\left[\frac14 - \frac{1}{(3n+1)(3n+4)}\right] = \frac{n(3n+5)}{8(3n+1)(3n+4)} \).

Teacher's Note:
1. Resolve each term into a difference of two successive products.
2. Check n = 1: 8/(8 x 4 x 7) = 1/28.

Question 24: Two sets of three positive terms in AP each have sum 15. The common difference of the first set exceeds that of the second by 1, and the ratio of the product of the first set to the product of the second is 7 : 8. The first set is:
(a) \( 2, 5, 8 \)
(b) \( 1, 5, 9 \)
(c) \( 3, 5, 7 \)
(d) \( 4, 5, 6 \)
Show Answer & Explanation

Answer: (c) \( 3, 5, 7 \)

Explanation:
1. Each set has middle term 5. For common differences \( d_1 = d_2 + 1 \): \( \frac{25 - d_1^2}{25 - d_2^2} = \frac78 \).
2. \( 8(25 - d_1^2) = 7(25 - (d_1 - 1)^2) \Rightarrow d_1^2 + 14 d_1 - 32 = 0 \Rightarrow d_1 = 2 \) (or -16, which gives negative terms).
3. First set 3, 5, 7 (product 105); second set 4, 5, 6 (product 120): 105 : 120 = 7 : 8.

Teacher's Note:
1. The middle term of a three term AP with sum 15 is 5.
2. Check the product ratio.

Question 25: If a, b, c, d, e are five positive numbers such that a, b, c are in AP, b, c, d are in GP and c, d, e are in HP, with a = 2 and e = 18, then b, c, d are:
(a) \( 4, 6, 9 \)
(b) \( 3, 4, 6 \)
(c) \( 5, 8, 13 \)
(d) \( 6, 10, 16 \)
Show Answer & Explanation

Answer: (a) \( 4, 6, 9 \)

Explanation:
1. \( c = 2b - a = 2b - 2 \), \( d = \frac{c^2}{b} \) and \( e = \frac{(2b - a)^2}{a} \) (from the HP condition).
2. \( 18 = \frac{(2b - 2)^2}{2} \Rightarrow (2b - 2)^2 = 36 \Rightarrow b = 4 \) (positive), giving \( c = 6,\ d = 9 \).
3. Check: 2, 4, 6 AP; 4, 6, 9 GP; 6, 9, 18 HP.

Teacher's Note:
1. Express c, d, e in terms of a and b.
2. 6, 9, 18 has reciprocals 1/6, 1/9, 1/18 in AP.

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