# CBSE Class 11th Math 9 – Sequences and Series MCQs

#### What is the smallest three-digit number in the A.P, with first term -3 and common difference 7?

The given A.P. has first term = -3 and common difference = 7. Let the smallest three-digit number in the A.P. be its n^{th} term. Then a + (n, -, 1)d¡Ý 100 ⟹- 3 + (n, -, 1)7¡Ý100 ⟹-3+7n-7¡Ý100 ⟹7n-10¡Ý100 ⟹7n¡Ý110 ⟹n¡Ý 110/7 ¡àn = 16 Hence the smallest three-digit number in the A.P. is the 16^{th}term, i.e., -3 + 15×7 = 102.

#### If the sum of the 5^{th} term and the 15^{th}term of an A.P. is 18, then which of the following statements is DEFINITELY TRUE?

Given that the sum of the 5^{th} term and the 15^{th}terms of the A.P. is 18. That is, (a + 4d) + (a + 14d) = 18 ⟹2a+18d=18 The sum to n terms of an A.P. is given by S_{n}=n/2(2a, +, (n, -, 1), d) S_{19}=19/2 (2a, +, (19, -, 1), d) That is, S_{19}=19/2(2a, +, 18d) = 19/2 × 18 = 171 Therefore, the sum of the first 19 terms of the A.P. is 171.

#### If the sum of the first 17 terms of an AP is 153, what is the 9^{th} term of that AP?

The formula to find the sum to n terms of an AP is: S_{n} = n/2(2a, +, (n, -, 1)d) where n is the number of terms, a is the first term, and d is the common difference. And the n^{th} term is: t_{n} = a + (n - 1)d) It is given that S_{17} = 153. That is, 17/2(2a, +, (17, -, 1)d) = 153 ⟹ 17/2(2a, +, 16, d) =153 ⟹17(a, +, 8, d) = 153 ⟹17 × t_{9 } = 153 ⟹ t_{9} = 153/17 = 9

#### The sum to n terms of a series is given by S_{n} =3n(n, -, 5)(n, +, 1), for n ≥ 5. The difference between the 13^{th} term and the 8^{th}term of the series is _____.

S_{13 }- S_{12} will give the 13^{th} term and S_{8} - S_{7} will give the 8^{th}term of the series. t_{13} = S_{13} - S_{12} = [3 × 13 × (13 - 5) × (13 + 1)] - [3 × 12 × (12 - 5) × (12 + 1)] = 4368 - 3276 = 1092 t_{8 }= S_{8} - S_{7 } = [3 × 8 × (8 - 5) × (8 + 1)] - [3 × 7 × (7 - 5) × (7 + 1)] = 648 - 336 = 312 Therefore, t_{13 }- t_{8} = 1092 - 312 = 780

#### If the 48^{th} term of an A.P. is 84 and its 84^{th} term is 48, which term of the A.P. is 36?

Given that the 48^{th} term of an A.P. is 84 and its 84^{th}term is 48. Let the first term of the A.P. be a and the common difference be d. Then, a + 47d = 84 and a + 83d = 48 Subtracting the first equation from the second, we get (a + 83d) - (a + 47d) = -36 ⟹ 36d = -36 d = -1 Hence, a = 84 - 47d = 84 + 47 = 131 Let the p^{th } term of this A.P. be 36. That is, 131 + (p - 1)×(-1) = 36 On simplification, this gives p = 96. Therefore, the 96^{th} term of the given A.P. is 36.

#### If two geometric means A and B are inserted between the numbers m and n, then A^{2 }+ B^{2} is equal to ____.

Let A = mr; B = mr^{2 }and n = mr^{3} Then, A^{2} + B^{2} = (mr)^{2} + (mr^{2})^{2} = m^{2}r^{2} + m^{2}r^{4} = m(mr^{2}) + (mr)(mr^{3}) = mB + nA

#### In an A.P., the first term is -1 and eleven times the sum of the first three terms is equal to twice the sum of the next three terms. Find the common difference of the A.P.

Let a be the first term of the A.P. and d be its common difference. Given that eleven times the sum of the first three terms is equal to twice the sum of the next three terms. That is, 11×[a + (a + d) + (a + 2d)] = 2×[(a + 3d) + (a + 4d) + (a + 5d)] ⟹ 11×[3a + 3d] = 2×[3a + 12d] ⟹ 33a + 33d = 6a + 24d ⟹ 9d = -27a ⟹ d = -3a Given that a = -1. d = 3

#### The sum to n terms of an A.P. is 1950. The first term of the A.P. is 72 and the nth term is 708. Find the common difference of the A.P.

The sum to n terms of an A.P. is 1950. The first term of the A.P. is 72 and the n^{th} term is 708. S_{n} = n/2 (First term +Last term) 1950 = n/2 (72+708) = 390n ⟹ n = 5 So the 5^{th} term is 708. That is, 72 + 4d = 708 ⟹ 4d = 708 - 72 = 636 d = 636/4 = 159

#### The 100^{th}term of the series 8 × 7 + 10 × 5 + 12 × 3 + ... is _______.

The given series comes from two arithmetic sequences: 8, 10, 12 ... and 7, 5, 3... Each term in the given series is the product of the corresponding terms in these sequences. The n^{th} term of the first AP is t_{n} = a + (n - 1)d = 8 + (n - 1) × 2 = 2n + 6 The n^{th} term of the second AP is t_{n} = a + (n - 1)d = 7 + (n - 1) × (-2) = -2n + 9 Therefore, the n^{th} term of the given series is (2n + 6)(-2n + 9) = -4n^{2}+ 6n + 54 Therefore, 100^{th}term = -39346.

#### In an AP, if the 5th and 12th terms are 30 and 65 respectively, then the sum of first 20 terms is ____.

Let the first term and common difference of AP be a and d. Given, a_{5 } = a + 4d = 30 - - - - (1) a_{12 } = a +11d = 65 - - - - (2) solving (1) and (2), we get a = 10 and d = 5 Sum of the first 20 terms = S_{20}=20/2[2×10+(20−1)5] = 10[20 + 95] = 10 × 115 = 1150

#### Two geometric means x and y are inserted between 1 and 4096. If a, b and c are three geometric means inserted between 1 and 4096, then which of the following is NOT TRUE?

x and y are two geometric means inserted between 1 and 4096. Let 1 × r = x and 1 × r^{2 }= y, where r^{3 } = 4096/1 Therefore, r = 16; x = 16 and y = 256 a, b and c are three geometric means inserted between 1 and 4096. Let 1 × R = a; 1 × R^{2} = b and 1 × R^{3} = c, where R^{4 } = 4096/1 R = 8 Therefore, a = 8; b = 64 and c = 512 Option A: xy = 16 × 256 = 4096; b^{2} = 64^{2}= 4096. Hence, TRUE. Option B: x < b < y implies 16 < 64 < 256. Hence, TRUE. Option C: y < a + b implies 256 x + y implies 512 > 16 + 256 = 272. Hence, TRUE.

#### If the difference between the 12^{th} and 7^{th}terms of an increasing A.P. is 63 more than the difference between its 6^{th} and 4^{th}terms, then the common difference.

Given that the difference between the 12^{th} and 7^{th}terms of an increasing A.P. is 63 more than the difference between its 6th and 4th terms. That is, [(a + 11d) - (a + 6d)] = 63 + [(a + 5d) - (a + 3d)] ⟹ 5d = 63 + 2d i.e. 3d = 63 d = 21

#### If the sum of the third term and the tenth term of an A.P. is 150, what is the sum of its first 12 terms?

Given that the sum of the third term and the tenth term of an A.P. is 150. That is, (a + 2d) + (a + 9d) = 150 ⟹ 2a + 11d = 150 Therefore, the sum of the first 12 terms is: S_{12} = 12/2 (2a+11d) = 6 × 150 = 900

#### The sum of 10 terms of an AP is 105 and the sum of 15 terms of the AP is 270. Find its 12^{th}term.

The formula to find the sum to n terms of an AP is: S_{n } = n/2(2a, +, (n, -, 1), d) where n is the number of terms, a is the first term, and d is the common difference. It is given that S_{10} = 105. That is, 10/2(2a, +, 9d) = 105 ∴ 5(2a, +, 9d) = 105 ⟹2a + 9d = 105/5 = 21⟶(I) Similarly, since it is given that S_{15} = 270, we have 15/2(2a, +, 14d) = 270 ∴ 15(a, +, 7d) = 270 ⟹ a + 7d = 270/15 = 18⟶ (II) 2×(II) - (I) gives d = 3. Substituting d = 3 in equation (II), a + 7 × 3 = 18 ⟹ a = 18 – 21 = -3 Therefore, the 12^{th} term is: a + 11d = (-3) + 11×3 = 30.

#### The sum of n terms of a series is given by S_{n}=n^{2}-5n. Which term of the series is 50?

Given that S_{n}=n^{2}-5n. If the p^{th }term is 50, then S_{p} - S_{p-1} = 50. ⟹( p^{2}, -, 5p) – ((p, -, 1)^{2} , -, 5, (p, -, 1) = 50 ⟹( p^{2}, ^{ }-, 5p) - )(p^{2}, -, 2p, +, 1, -, 5p, +, 5) = 50 ⟹2p-6=50 ⟹2p=56 ⟹p=28 Therefore, the 28^{th}term of the series is 50.