The third term of a G.P. is 4. The product of the first five terms is:
Answer: B
Let the first five terms of the G.P. be \(a/r^2, a/r, a, ar, ar^2\). The third term is \(a=4\). The product of these five terms is \((a/r^2) \times (a/r) \times a \times (ar) \times (ar^2) = a^5\). Since a=4, the product is \(4^5 = 1024\).
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The sum of the first n terms of an A.P. is given by \(S_n = 3n^2 + 5n\). Find the common difference.
Answer: B
The nth term of an A.P. can be found by \(a_n = S_n - S_{n-1}\). First term \(a_1 = S_1 = 3(1)^2 + 5(1) = 8\). Sum of first two terms \(S_2 = 3(2)^2 + 5(2) = 12 + 10 = 22\). The second term \(a_2 = S_2 - S_1 = 22 - 8 = 14\). The common difference \(d = a_2 - a_1 = 14 - 8 = 6\). Alternatively, for a sum in the form \(An^2+Bn\), the common difference is \(2A\). Here A=3, so \(d=2 \times 3=6\).
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The sum of three numbers in an A.P. is 27 and their product is 504. Find the numbers.
Answer: A
Let the three numbers be \(a-d, a, a+d\). Their sum is \((a-d)+a+(a+d) = 3a = 27\), so \(a=9\). Their product is \((a-d)a(a+d) = a(a^2-d^2) = 504\). Substituting a=9, we get \(9(81-d^2)=504\). \(81-d^2 = 504/9 = 56\). \(d^2 = 81-56 = 25\), so \(d = \pm 5\). If d=5, the numbers are 4, 9, 14. If d=-5, the numbers are 14, 9, 4. The numbers are 4, 9, 14.
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The 4th term of an H.P. is 1/3 and the 7th term is 1/4. Find the 10th term.
Answer: A
If the terms are in H.P., their reciprocals are in A.P. The 4th term of the A.P. is 3 and the 7th term is 4. Let the A.P. be \(a, a+d, ...\). \(a_4 = a+3d = 3\) and \(a_7 = a+6d = 4\). Subtracting the equations gives \(3d=1\), so \(d=1/3\). Substituting back, \(a + 3(1/3) = 3 \Rightarrow a+1=3 \Rightarrow a=2\). The 10th term of the A.P. is \(a_{10} = a+9d = 2 + 9(1/3) = 2+3=5\). The 10th term of the H.P. is the reciprocal, which is 1/5.
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The sum of the series \(1 - 2 + 3 - 4 + ...\) up to 100 terms is:
Answer: B
We can group the terms in pairs: \((1-2) + (3-4) + (5-6) + ... + (99-100)\). Each pair sums to -1. Since there are 100 terms, there are \(100/2 = 50\) such pairs. The total sum is \(50 \times (-1) = -50\).
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The relationship between Arithmetic Mean (A), Geometric Mean (G) and Harmonic Mean (H) is:
Answer: B
For any two positive numbers, the square of the Geometric Mean is equal to the product of the Arithmetic Mean and the Harmonic Mean. This is a standard property: \(G^2 = AH\).
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The sum of the first 'n' terms of an A.P. is 288. If the first term is 2 and the common difference is 4, find 'n'.
Answer: B
We use the formula for the sum of an A.P.: \(S_n = \frac{n}{2}[2a + (n-1)d]\).
Given \(S_n=288, a=2, d=4\).
\(288 = \frac{n}{2}[2(2) + (n-1)4]\)
\(288 = \frac{n}{2}[4 + 4n - 4]\)
\(288 = \frac{n}{2}[4n] = 2n^2\)
\(n^2 = 288/2 = 144\)
\(n = \sqrt{144} = 12\).
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Find the sum of the series: \(1^2 + 2^2 + 3^2 + ... + 10^2\).
Answer: B
The sum of the squares of the first n natural numbers is given by the formula \(S_n = \frac{n(n+1)(2n+1)}{6}\). Here, n=10. \(S_{10} = \frac{10(10+1)(2 \times 10+1)}{6} = \frac{10 \times 11 \times 21}{6} = 5 \times 11 \times 7 = 385\).
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Find the sum of the first 20 terms of the A.P. 1, 4, 7, 10, ...
Answer: A
Here, \(a = 1\), \(d = 3\), and \(n = 20\). The formula for the sum of the first n terms of an A.P. is \(S_n = \frac{n}{2}[2a + (n-1)d]\). So, \(S_{20} = \frac{20}{2}[2(1) + (20-1)3] = 10[2 + 19 \times 3] = 10[2+57] = 10 \times 59 = 590\).
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If the terms \(a, b, c\) are in Harmonic Progression (H.P.), then their reciprocals \(1/a, 1/b, 1/c\) are in...
Answer: C
By definition, a sequence of numbers is in Harmonic Progression if the reciprocals of the terms are in Arithmetic Progression. Therefore, if a, b, and c are in H.P., then 1/a, 1/b, and 1/c are in A.P.
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