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 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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The sum of all two-digit numbers divisible by 3 is:
Answer: A
The two-digit numbers divisible by 3 are 12, 15, ..., 99. This is an A.P. with \(a=12, d=3, a_n=99\). First, find the number of terms: \(a_n = a+(n-1)d \Rightarrow 99 = 12+(n-1)3 \Rightarrow 87 = (n-1)3 \Rightarrow 29 = n-1 \Rightarrow n=30\). The sum is \(S_n = \frac{n}{2}(a+a_n) = \frac{30}{2}(12+99) = 15(111) = 1665\).
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How many terms are there in the G.P. 3, 6, 12, ..., 384?
Answer: B
Here, \(a=3, r=2, a_n=384\). We use the formula \(a_n = ar^{n-1}\). \(384 = 3 \times 2^{n-1} \Rightarrow 128 = 2^{n-1}\). Since \(128 = 2^7\), we have \(2^7 = 2^{n-1}\). Equating the powers gives \(7=n-1\), so \(n=8\). There are 8 terms.
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If the second term of an H.P. is 1/5 and the 6th term is 1/13, what is the 10th term?
Answer: B
The corresponding A.P. has a 2nd term of 5 and a 6th term of 13. So, \(a+d=5\) and \(a+5d=13\). Subtracting gives \(4d=8\), so \(d=2\). Substituting d=2 gives \(a+2=5\), so \(a=3\). The 10th term of the A.P. is \(a_{10} = a+9d = 3+9(2) = 3+18=21\). The 10th term of the H.P. is the reciprocal, 1/21.
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The sum of an infinite geometric series is 15 and the sum of the squares of these terms is 45. Find the series.
Answer: A
Let the series be \(a, ar, ar^2, ...\). Sum \(\frac{a}{1-r} = 15\). The series of squares is \(a^2, a^2r^2, a^2r^4, ...\) which is a G.P. with first term \(a^2\) and common ratio \(r^2\). Its sum is \(\frac{a^2}{1-r^2} = 45\). We can write this as \(\frac{a^2}{(1-r)(1+r)} = 45\). Substituting \(a=15(1-r)\) is complex. Let's use \(\frac{a}{1-r} \times \frac{a}{1+r} = 45 \Rightarrow 15 \times \frac{a}{1+r} = 45 \Rightarrow \frac{a}{1+r} = 3 \Rightarrow a = 3(1+r)\). Now we have \(15(1-r) = 3(1+r) \Rightarrow 5(1-r)=1+r \Rightarrow 5-5r=1+r \Rightarrow 4=6r \Rightarrow r=2/3\). Then \(a = 15(1-2/3) = 15(1/3)=5\). The series is 5, 10/3, 20/9, ...
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The angles of a quadrilateral are in A.P. with a common difference of 10 degrees. Find the smallest angle.
Answer: B
Let the angles be \(a, a+10, a+20, a+30\). The sum of angles in a quadrilateral is 360°. So, \(a + (a+10) + (a+20) + (a+30) = 360 \Rightarrow 4a + 60 = 360 \Rightarrow 4a = 300 \Rightarrow a = 75\). The smallest angle is 75°.
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The geometric mean between two numbers is 12. If one number is 9, what is the other number?
Answer: B
Let the numbers be a and b. G.M. = \(\sqrt{ab} = 12\), so \(ab = 144\). Given one number a=9, we have \(9b = 144\), so \(b = 144/9 = 16\). The other number is 16.
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Insert 4 arithmetic means between 4 and 19.
Answer: A
We need to form an A.P. with 6 terms, where the first term is 4 and the 6th term is 19. So, \(a=4\) and \(a_6=19\). \(a_6 = a+5d \Rightarrow 19=4+5d \Rightarrow 15=5d \Rightarrow d=3\). The means are the 2nd, 3rd, 4th, and 5th terms: \(a+d=7\), \(a+2d=10\), \(a+3d=13\), \(a+4d=16\).
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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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