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?

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x and y are two geometric means inserted between 1 and 4096. Let 1 × r = x and 1 × r2 = y, where r3 = 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 × R2 = b and 1 × R3 = c, where R4 = 4096/1 R = 8 Therefore, a = 8; b = 64 and c = 512 Option A: xy = 16 × 256 = 4096; b2 = 642= 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 sum of the first 17 terms of an AP is 153, what is the 9th term of that AP?

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The formula to find the sum to n terms of an AP is: Sn = 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 nth term is: tn = a + (n - 1)d) It is given that S17 = 153. That is, 17/2(2a, +, (17, -, 1)d) = 153 ⟹ 17/2(2a, +, 16, d) =153 ⟹17(a, +, 8, d) = 153 ⟹17 × t9 = 153 ⟹ t9 = 153/17 = 9

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.

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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. ˆ d = 3

The sum of 10 terms of an AP is 105 and the sum of 15 terms of the AP is 270. Find its 12thterm.

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The formula to find the sum to n terms of an AP is: Sn = 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 S10 = 105. That is, 10/2(2a, +, 9d) = 105 ∴ 5(2a, +, 9d) = 105 ⟹2a + 9d = 105/5 = 21⟶(I) Similarly, since it is given that S15 = 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 12th term is: a + 11d = (-3) + 11×3 = 30.

The 100thterm of the series 8 × 7 + 10 × 5 + 12 × 3 + ... is _______.

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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 nth term of the first AP is tn = a + (n - 1)d = 8 + (n - 1) × 2 = 2n + 6 The nth term of the second AP is tn = a + (n - 1)d = 7 + (n - 1) × (-2) = -2n + 9 Therefore, the nth term of the given series is (2n + 6)(-2n + 9) = -4n2+ 6n + 54 Therefore, 100thterm = -39346.

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

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Given that the difference between the 12th and 7thterms 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 ˆ d = 21

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

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The given A.P. has first term = -3 and common difference = 7. Let the smallest three-digit number in the A.P. be its nth 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 16thterm, i.e., -3 + 15×7 = 102.

The sum of n terms of a series is given by Sn=n2-5n. Which term of the series is 50?

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Given that Sn=n2-5n. If the pth term is 50, then Sp - Sp-1 = 50. ⟹( p2, -, 5p) – ((p, -, 1)2 , -, 5, (p, -, 1) = 50 ⟹( p2, -, 5p) - )(p2, -, 2p, +, 1, -, 5p, +, 5) = 50 ⟹2p-6=50 ⟹2p=56 ⟹p=28 Therefore, the 28thterm of the series is 50.

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.

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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. Sn = n/2 (First term +Last term) ˆ 1950 = n/2 (72+708) = 390n ⟹ n = 5 So the 5th term is 708. That is, 72 + 4d = 708 ⟹ 4d = 708 - 72 = 636 d = 636/4 = 159

If two geometric means A and B are inserted between the numbers m and n, then A2 + B2 is equal to ____.

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Let A = mr; B = mr2 and n = mr3 Then, A2 + B2 = (mr)2 + (mr2)2 = m2r2 + m2r4 = m(mr2) + (mr)(mr3) = mB + nA

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?

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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: S12 = 12/2 (2a+11d) = 6 × 150 = 900

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

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Given that the 48th term of an A.P. is 84 and its 84thterm 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 ˆ d = -1 Hence, a = 84 - 47d = 84 + 47 = 131 Let the pth term of this A.P. be 36. That is, 131 + (p - 1)×(-1) = 36 On simplification, this gives p = 96. Therefore, the 96th term of the given A.P. is 36.

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

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Given that the sum of the 5th term and the 15thterms 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 Sn=n/2(2a, +, (n, -, 1), d)  S19=19/2 (2a, +, (19, -, 1), d) That is, S19=19/2(2a, +, 18d) = 19/2 × 18 = 171 Therefore, the sum of the first 19 terms of the A.P. is 171.

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

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Let the first term and common difference of AP be a and d. Given, a5 = a + 4d = 30 - - - - (1) a12 = a +11d = 65 - - - - (2) solving (1) and (2), we get a = 10 and d = 5 Sum of the first 20 terms = S20=20/2[2×10+(20−1)5] = 10[20 + 95] = 10 × 115 = 1150

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

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S13 - S12 will give the 13th term and S8 - S7 will give the 8thterm of the series. t13 = S13 - S12 = [3 × 13 × (13 - 5) × (13 + 1)] - [3 × 12 × (12 - 5) × (12 + 1)] = 4368 - 3276 = 1092 t8 = S8 - S7 = [3 × 8 × (8 - 5) × (8 + 1)] - [3 × 7 × (7 - 5) × (7 + 1)] = 648 - 336 = 312 Therefore, t13 - t8 = 1092 - 312 = 780

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CBSE Class 11th Math 9 - Sequences and Series MCQs
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