JEE Mathematics Sequence and Series MCQs Set 09

Answer: (a) 13

Explanation:
1. The two-digit multiples of 7 are 14, 21, ..., 98.
2. Using \( a_n = a + (n-1)d \): \( 98 = 14 + (n-1)7 \), so n = 13.

Teacher's Note:
1. List the first and last multiples, then use the nth term formula.

Answer: (b) 108

Explanation:
1. The first multiple of 6 above 103 is 108 and the last below 750 is 750 itself (750 = 6 × 125).
2. \( 750 = 108 + (n-1)6 \), so n − 1 = 107 and n = 108.

Teacher's Note:
1. Find the first and last multiples inside the range carefully.
2. Then count with the nth term formula.

Answer: (a) 1830

Explanation:
1. The natural numbers form an AP with a = 1, d = 1.
2. \( S_{60} = \frac{60 \times 61}{2} = 1830 \).

Teacher's Note:
1. Use \( \frac{n(n+1)}{2} \) for the sum of the first n natural numbers.

Answer: (b) 4905

Explanation:
1. Two-digit numbers 10, 11, ..., 99 form an AP with 90 terms.
2. \( S = \frac{90}{2}(10 + 99) = 45 \times 109 = 4905 \).

Teacher's Note:
1. First term 10, last term 99, number of terms 90.
2. Use \( S = \frac{n}{2}(a + l) \).

Answer: (c) −37

Explanation:
1. Here a = 7 and d = −2.
2. \( a_{23} = 7 + 22 \times (-2) = 7 - 44 = -37 \).

Teacher's Note:
1. The common difference is negative because the terms decrease.

Answer: (a) 3

Explanation:
1. In an AP the middle term is the mean of its neighbours: \( 2 \times 3k = (k+1) + (4k+2) \).
2. \( 6k = 5k + 3 \), so k = 3.

Teacher's Note:
1. Three terms a, b, c are in AP when 2b = a + c.

Answer: (c) 106th

Explanation:
1. Here a = 5 and d = 3.
2. \( 320 = 5 + (n-1)3 \) gives \( n - 1 = 105 \), so n = 106.

Teacher's Note:
1. Solve \( a + (n-1)d = \) the given term for n.

Answer: (b) 9

Explanation:
1. \( a + 4d = 5 \) and \( a + 12d = -3 \).
2. Subtracting: \( 8d = -8 \), so d = −1.
3. \( a = 5 + 4 = 9 \).

Teacher's Note:
1. Two terms give two linear equations in a and d.

Answer: (b) 17th

Explanation:
1. a = 64 and d = −4.
2. \( 64 + (n-1)(-4) = 0 \) gives \( n - 1 = 16 \), so n = 17.

Teacher's Note:
1. Set the nth term equal to zero and solve.

Answer: (a) 26

Explanation:
1. Put n = 7: \( 3 \times 7 + 5 = 26 \).

Teacher's Note:
1. The nth term formula can be used directly by substituting n.

Answer: (c) 3 : 4 : 5

Explanation:
1. Let the sides be a − d, a, a + d.
2. Pythagoras: \( (a-d)^2 + a^2 = (a+d)^2 \) gives \( a^2 = 4ad \), so a = 4d.
3. The sides are 3d, 4d, 5d, in the ratio 3 : 4 : 5.

Teacher's Note:
1. Three numbers in AP are best taken as a − d, a, a + d.
2. The 3 : 4 : 5 triangle is the only right triangle with sides in AP.

Answer: (a) 400

Explanation:
1. The sum of the first n odd numbers is \( n^2 \).
2. For n = 20 the sum is 400. Check: \( \frac{20}{2}(2 + 19 \times 2) = 400 \).

Teacher's Note:
1. The sum of the first n odd numbers equals \( n^2 \).

Answer: (a) 2139

Explanation:
1. a = 5, d = 8 and l = 181.
2. \( 181 = 5 + (n-1)8 \) gives n = 23.
3. \( S = \frac{23}{2}(5 + 181) = 23 \times 93 = 2139 \).

Teacher's Note:
1. Find the number of terms first, then use \( S = \frac{n}{2}(a+l) \).

Answer: (c) 7500

Explanation:
1. The odd numbers are 101, 103, ..., 199, so a = 101, l = 199 and d = 2.
2. \( 199 = 101 + (n-1)2 \) gives n = 50.
3. \( S = \frac{50}{2}(101 + 199) = 7500 \).

Teacher's Note:
1. Between 100 and 200 excludes both ends, but neither is odd.
2. Count the terms before summing.

Answer: (a) 950

Explanation:
1. The multiples are 5, 10, ..., 95, so n = 19.
2. \( S = \frac{19}{2}(5 + 95) = 950 \).

Teacher's Note:
1. Less than 100 excludes 100 itself.
2. The last multiple is 95.

Answer: (a) 2450

Explanation:
1. The numbers are 2, 4, ..., 98, so n = 49.
2. \( S = \frac{49}{2}(2 + 98) = 2450 \).

Teacher's Note:
1. Less than 100 means the last even number is 98.
2. There are 49 terms.

Answer: (a) 16

Explanation:
1. The arithmetic mean of two numbers is half their sum.
2. \( \frac{14 + 18}{2} = 16 \).

Teacher's Note:
1. The single A.M. between a and b is \( \frac{a+b}{2} \).

Answer: (b) 22

Explanation:
1. The first term is 4 and the fifth term is 28, so \( 4d = 24 \) and d = 6.
2. \( A_3 \) is the fourth term: \( 4 + 3 \times 6 = 22 \).

Teacher's Note:
1. The fifth term fixes the common difference.
2. A₃ is the fourth term, not the third.

Answer: (c) 8

Explanation:
1. \( S_n = \frac{n}{2}[6 + (n-1)3] = \frac{3n(n+1)}{2} = 108 \).
2. \( n(n+1) = 72 \), so n = 8.

Teacher's Note:
1. Form an equation in n from the sum formula.
2. The negative root is rejected.

Answer: (b) −8

Explanation:
1. \( a_8 - a_6 = 2d = 10 \), so d = 5.
2. \( a_2 = a_6 - 4d = 12 - 20 = -8 \).

Teacher's Note:
1. Subtract the two terms to find d quickly.

Answer: (b) 76

Explanation:
1. \( a_{25} = S_{25} - S_{24} \).
2. \( S_{25} = \frac{3 \times 625 + 125}{2} = 1000 \) and \( S_{24} = \frac{3 \times 576 + 120}{2} = 924 \).
3. \( a_{25} = 1000 - 924 = 76 \).

Teacher's Note:
1. The nth term is the difference of consecutive sums, \( a_n = S_n - S_{n-1} \).

Answer: (b) 16, 5

Explanation:
1. The rows form an AP with a = 20 and d = −1.
2. \( S_n = \frac{n}{2}[40 - (n-1)] = 200 \) gives \( n^2 - 41n + 400 = 0 \), so n = 16 or n = 25.
3. For n = 25 the top row would have −4 logs, which is impossible, so n = 16 and the top row has \( 20 - 15 = 5 \) logs.

Teacher's Note:
1. Form the equation from the sum formula.
2. Reject the root that gives a negative number of logs.