Find the sum of the first 10 terms of the H.P. \(1/2, 1/5, 1/8, ...\)
Answer: D
There is no direct formula to find the sum of a Harmonic Progression. We can find the sum of the corresponding A.P. (2, 5, 8, ...), but this does not give the sum of the H.P. The sum of an H.P. cannot be calculated using a simple standard formula.
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Find the 10th term of the Arithmetic Progression (A.P.) 2, 7, 12, ...
Answer: A
In this A.P., the first term \(a = 2\) and the common difference \(d = 7 - 2 = 5\). The formula for the nth term is \(a_n = a + (n-1)d\). For the 10th term (n=10), we have \(a_{10} = 2 + (10-1) \times 5 = 2 + 9 \times 5 = 2 + 45 = 47\).
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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 fourth term of an A.P. is 11 and the seventh term is 20. Find the 15th term.
Answer: B
We have \(a+3d=11\) and \(a+6d=20\). Subtracting the equations gives \(3d=9\), so \(d=3\). Substituting back, \(a+3(3)=11 \Rightarrow a+9=11 \Rightarrow a=2\). The 15th term is \(a_{15} = a+14d = 2 + 14(3) = 2+42=44\).
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The sum of the series \(1 + 2x + 3x^2 + 4x^3 + ...\) for \(|x|<1\) is:
Answer: C
This is an Arithmetico-Geometric Progression (A.G.P.) with the A.P. part being 1, 2, 3, ... and G.P. part being \(1, x, x^2, ...\). The sum of an infinite A.G.P. is given by \(S = \frac{a}{1-r} + \frac{dr}{(1-r)^2}\). Here, the first term of the A.P. is \(a_{AP}=1\), \(d=1\), and \(r=x\). The first term of the A.G.P. is \(a=1\). The sum is \(S = \frac{1}{1-x} + \frac{1 \cdot x}{(1-x)^2} = \frac{1-x+x}{(1-x)^2} = \frac{1}{(1-x)^2}\). This is a standard result for the expansion of \((1-x)^{-2}\).
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The first term of a G.P. is 1. The sum of the third and fifth term is 90. Find the common ratio.
Answer: B
Given \(a=1\). The third term is \(ar^2 = r^2\) and the fifth term is \(ar^4 = r^4\). Given \(r^2 + r^4 = 90\). Let \(x=r^2\). The equation is \(x+x^2=90 \Rightarrow x^2+x-90=0 \Rightarrow (x+10)(x-9)=0\). Since \(x=r^2\), it cannot be negative. So \(x=9\). \(r^2=9\), which gives \(r = \pm 3\).
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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 first three terms of a G.P. is to the sum of first six terms as 125:152. Find the common ratio.
Answer: B
We have \(\frac{S_3}{S_6} = \frac{125}{152}\). Let the sum formula be \(S_n=a(r^n-1)/(r-1)\). So, \(\frac{a(r^3-1)/(r-1)}{a(r^6-1)/(r-1)} = \frac{r^3-1}{r^6-1} = \frac{r^3-1}{(r^3-1)(r^3+1)} = \frac{1}{r^3+1}\). We have \(\frac{1}{r^3+1} = \frac{125}{152} \Rightarrow r^3+1 = \frac{152}{125} \Rightarrow r^3 = \frac{152}{125} - 1 = \frac{27}{125}\). So \(r = 3/5\).
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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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