Progressions (A.P., G.P., H.P.) Questions and Answers

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.

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, \(S_{20} = \frac{20}{2}[2(1) + (20-1)3] = 10[2 + 19 \times 3] = 10[2+57] = 10 \times 59 = 590\).

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 \(a_5 = 5 \times 2^{5-1} = 5 \times 2^4 = 5 \times 16 = 80\).

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.

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, ...

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. \(S_\infty = \frac{16}{1 - 1/2} = \frac{16}{1/2} = 32\).

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.

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.

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 \(d=2 \times 3=6\).

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 \(4^5 = 1024\).