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
Divide any term by the one before it.
\[ \frac{6}{3} = \frac{12}{6} = \frac{24}{12} = 2 \]
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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
A harmonic progression is defined as a sequence whose reciprocals form an arithmetic progression.
\[ a, b, c \text{ in H.P.} \iff \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \text{ in A.P.} \]
This is why H.P. problems are almost always solved by inverting first.
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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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What is the 5th term of the G.P. 5, 10, 20, ...?
Answer: C
In this G.P., the first term \(a = 5\) and the common ratio \(r = 10/5 = 2\). The formula for the nth term is \(a_n = ar^{n-1}\). For the 5th term (n=5), we have \(a_5 = 5 \times 2^{5-1} = 5 \times 2^4 = 5 \times 16 = 80\).
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Find the Harmonic Mean (H.M.) between 4 and 6.
Answer: B
\[ \text{H.M.} = \frac{2ab}{a+b} = \frac{2 \times 4 \times 6}{10} \]
\[ = \frac{48}{10} = 4.8 \]
The harmonic mean always falls below the arithmetic mean of 5.
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The 7th term of an A.P. is 20 and the 13th term is 32. Find the A.P.
Answer: A
Let the first term be 'a' and the common difference be 'd'. We have \(a_7 = a + 6d = 20\) and \(a_{13} = a + 12d = 32\). Subtracting the first equation from the second gives \(6d = 12\), so \(d=2\). Substituting d=2 into the first equation gives \(a + 6(2) = 20\), so \(a + 12 = 20\) and \(a=8\). The A.P. is 8, 10, 12, ...
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Find the sum of the infinite G.P. 16, 8, 4, ...
Answer: A
The first term \(a = 16\) and the common ratio \(r = 8/16 = 1/2\). The sum of an infinite G.P. is given by \(S_\infty = \frac{a}{1-r}\), provided \(|r| < 1\). Here, \(|1/2| < 1\), so the sum exists. \(S_\infty = \frac{16}{1 - 1/2} = \frac{16}{1/2} = 32\).
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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 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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