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

Answer: B

The fourth term is the difference of the two sums.

\[ 20 - 12 = 8 \]

No property of the progression is needed - only that the sums are of consecutive terms.

Answer: B

An infinite geometric series with \( |r| < 1 \) sums to

\[ S = \frac{a}{1-r} = \frac{1}{1 - 1/3} = \frac{1}{2/3} = \frac{3}{2} \]

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 \(a_{15} = a+14d = 2 + 14(3) = 2+42=44\).

Answer: A

If the terms are in G.P., the ratio of consecutive terms is constant. So, \(\frac{2x+2}{x} = \frac{3x+3}{2x+2}\). \((2x+2)^2 = x(3x+3) \Rightarrow 4x^2+8x+4 = 3x^2+3x \Rightarrow x^2+5x+4=0 \Rightarrow (x+1)(x+4)=0\). So \(x=-1\) or \(x=-4\). If x=-1, terms are -1, 0, 0 (not a G.P.). If x=-4, terms are -4, -6, -9. This is a G.P. with \(r = -6/-4 = 3/2\). The 4th term is \(-9 \times (3/2) = -27/2 = -13.5\).

Answer: D

Let the terms be \(a, ar, ar^2\). Given \(a+ar^2=20 \Rightarrow a(1+r^2)=20\). And \(a+ar+ar^2=26 \Rightarrow a(1+r+r^2)=26\). Dividing the second eq by the first: \(\frac{1+r+r^2}{1+r^2} = \frac{26}{20} = \frac{13}{10}\). \(10+10r+10r^2 = 13+13r^2 \Rightarrow 3r^2-10r+3=0 \Rightarrow (3r-1)(r-3)=0\). So \(r=3\) or \(r=1/3\).

Answer: A

The three-digit numbers are from 100 to 999. First three-digit number divisible by 7 is 105. Last is 994. This is an A.P. with \(a=105, d=7, a_n=994\). We use \(a_n = a+(n-1)d\). \(994 = 105+(n-1)7 \Rightarrow 889 = (n-1)7 \Rightarrow n-1 = 127 \Rightarrow n=128\). There are 128 such numbers.

Answer: A

Let the numbers be \(a/r, a, ar\). Product is \((a/r)(a)(ar) = a^3 = 216\), so \(a=6\). Sum of products in pairs: \((a/r)(a) + a(ar) + (a/r)(ar) = a^2/r + a^2r + a^2 = 156\). Substitute a=6: \(36/r + 36r + 36 = 156 \Rightarrow 36/r+36r=120\). Divide by 12: \(3/r+3r=10 \Rightarrow 3+3r^2=10r \Rightarrow 3r^2-10r+3=0 \Rightarrow (3r-1)(r-3)=0\). So \(r=3\) or \(r=1/3\). If r=3, numbers are 2, 6, 18. If r=1/3, numbers are 18, 6, 2.

Answer: A

The numbers are 108, 117, ..., 999. This is an A.P. with \(a=108, d=9, a_n=999\). Number of terms: \(999=108+(n-1)9 \Rightarrow 891=(n-1)9 \Rightarrow n-1=99 \Rightarrow n=100\). Sum = \(\frac{100}{2}(108+999) = 50(1107) = 55350\).

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.