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

Answer: C

Correct answer: Rs. 18600

The monthly savings form an arithmetic progression with \(a = 200\) and \(d = 50\), and two years is \(n = 24\) months.

\[S_n = \frac{n}{2}\left[2a + (n-1)d\right]\]

\[S_{24} = \frac{24}{2}\left[2(200) + 23(50)\right] = 12\left[400 + 1150\right] = 12 \times 1550 = 18600\]

So he will have saved Rs. 18600.

Answer: B

The savings form an Arithmetic Progression with the first term \(a = 200\) and common difference \(d = 50\). We need to find the sum for 2 years, which is \(n=24\) months.

The formula for the sum is \(S_n = \frac{n}{2}[2a + (n-1)d]\).

\(S_{24} = \frac{24}{2}[2(200) + (24-1)50] = 12[400 + (23)(50)] = 12[400 + 1150] = 12(1550) = 18600\).

He will have saved Rs. 18,600 after 2 years.

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 \(r = \pm 3\).

Answer: C

The corresponding A.P. has 5th term 1/7 and 7th term 1/5. So, \(a+4d=1/7\) and \(a+6d=1/5\). Subtracting gives \(2d = 1/5 - 1/7 = (7-5)/35 = 2/35\), so \(d=1/35\). Substituting back: \(a+4/35=1/7 \Rightarrow a = 1/7 - 4/35 = (5-4)/35 = 1/35\). The 12th term of the A.P. is \(a_{12} = a+11d = 1/35 + 11/35 = 12/35\). The 12th term of the H.P. is the reciprocal, 35/12.

Answer: B

Group the 100 terms into 50 consecutive pairs.

\[ (1-2) + (3-4) + \cdots + (99-100) \]

\[ = 50 \times (-1) = -50 \]

With an odd number of terms the last one would be unpaired and the sum positive.

Answer: B

To find the nth term from the end, we can use the formula \(l - (n-1)d\), where l is the last term. Here, \(l=49\), \(n=4\), and \(d = -8 - (-11) = 3\). The 4th term from the end is \(49 - (4-1) \times 3 = 49 - 9 = 40\). Alternatively, we can reverse the A.P. to 49, 46, ..., -11. Here \(a=49, d=-3\). The 4th term is \(49 + (4-1)(-3) = 49 - 9 = 40\).

Answer: A

Given \(a=5\) and \(a_3 = a+2d = 15\). So, \(5+2d=15 \Rightarrow 2d=10 \Rightarrow d=5\). We need the sum of the first 16 terms. \(S_{16} = \frac{16}{2}[2(5) + (16-1)5] = 8[10 + 15 \times 5] = 8[10+75] = 8(85) = 680\).

Answer: C

This is a trick question. The answer should be 600 m. Let's re-calculate. Initial drop distance = 120m. After that, it rebounds and falls. Rebound 1: up \(120 \times 2/3 = 80\), down 80. Rebound 2: up \(80 \times 2/3 = 160/3\), down 160/3. The total distance is \(120 + 2 \times (80 + 160/3 + ...)\). The series in the bracket is a G.P. with \(a=80, r=2/3\). Its sum is \(\frac{a}{1-r} = \frac{80}{1-2/3} = \frac{80}{1/3} = 240\). Total distance = \(120 + 2 \times 240 = 120 + 480 = 600\) m.

Answer: B

This is an A.P. with first term \(a=19\) and common difference \(d = 18.2 - 19 = -0.8\). We want to find the smallest n for which the nth term is less than 0. \(a_n = a + (n-1)d < 0 \Rightarrow 19 + (n-1)(-0.8) < 0 \Rightarrow 19 < 0.8(n-1) \Rightarrow 19/0.8 < n-1 \Rightarrow 23.75 < n-1 \Rightarrow 24.75 < n\). The smallest integer n satisfying this is 25. So, the 25th term is the first negative term.

Answer: B

Let the numbers be \(a/r, a, ar\). Their product is \(a^3 = 64\), so \(a=4\). Their sum is \(a/r + a + ar = 14 \Rightarrow 4/r + 4 + 4r = 14 \Rightarrow 4/r + 4r = 10 \Rightarrow 4+4r^2=10r \Rightarrow 4r^2-10r+4=0 \Rightarrow 2r^2-5r+2=0 \Rightarrow (2r-1)(r-2)=0\). So \(r=2\) or \(r=1/2\). If r=2, numbers are 2, 4, 8. If r=1/2, numbers are 8, 4, 2.