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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What is the common ratio of the Geometric Progression (G.P.) 3, 6, 12, 24, ...?
Answer: B
The common ratio (r) of a G.P. is found by dividing any term by its preceding term. \(r = \frac{6}{3} = 2\) or \(r = \frac{12}{6} = 2\). The common ratio is 2.
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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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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 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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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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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 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 first 9 terms of an A.P. is 81 and sum of first 18 terms is 324. Find the sum of first 27 terms.
Answer: A
Given \(S_9=81\) and \(S_{18}=324\). Notice that \(81=9^2 \times 1\) and \(324=18^2 \times 1\). No, that's not right. Let's use the formula. \(\frac{9}{2}(2a+8d)=81 \Rightarrow 2a+8d=18\). \(\frac{18}{2}(2a+17d)=324 \Rightarrow 2a+17d=36\). Subtracting the two equations gives \(9d=18 \Rightarrow d=2\). Then \(2a+16=18 \Rightarrow 2a=2 \Rightarrow a=1\). We need \(S_{27} = \frac{27}{2}[2(1)+(26)2] = \frac{27}{2}[2+52] = \frac{27}{2}(54) = 27 \times 27 = 729\).
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