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

Q: The sum of the series \(1 + 2x + 3x^2 + 4x^3 + ...\) for \(|x|

This is an Arithmetico-Geometric Progression (A.G.P.) with the A.P. part being 1, 2, 3, ... and G.P. part being \(1, x, x^2, ...\). The sum of an infinite A.G.P. is given by \(S = \frac{a}{1-r} + \frac{dr}{(1-r)^2}\). Here, the first term of the A.P. is \(a_{AP}=1\), \(d=1\), and \(r=x\). The first term of the A.G.P. is \(a=1\). The sum is \(S = \frac{1}{1-x} + \frac{1 \cdot x}{(1-x)^2} = \frac{1-x+x}{(1-x)^2} = \frac{1}{(1-x)^2}\). This is a standard result for the expansion of \((1-x)^{-2}\).

Q: What term of the G.P. \(2, 1, 1/2, 1/4, ...\) is 1/128?

Here, \(a=2, r=1/2, a_n=1/128\). Using the formula \(a_n = ar^{n-1}\): \(1/128 = 2 \times (1/2)^{n-1} \Rightarrow 1/256 = (1/2)^{n-1}\). Since \(256 = 2^8\), we have \((1/2)^8 = (1/2)^{n-1}\). Equating powers: \(8=n-1 \Rightarrow n=9\). It is the 9th term.

Q: 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.

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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81.

Find the sum of the series: \(5 + 55 + 555 + ...\) up to n terms.

A.\(\frac{5}{81}(10^{n+1}-9n-10)\)
B.\(\frac{5}{9}(10^n-1)\)
C.\(\frac{5}{81}(10^n-9n-1)\)
D.\(\frac{5}{9}(10^{n+1}-1)\)

Answer: A

The sum S can be written as \(S = 5(1+11+111+...)\). Multiply and divide by 9: \(S = \frac{5}{9}(9+99+999+...)\). This can be written as \(S = \frac{5}{9}[(10-1)+(100-1)+(1000-1)+...]\). \(S = \frac{5}{9}[(10+100+1000+...) - (1+1+...+n \text{ times})]\). The first part is a G.P. Its sum is \(10\frac{10^n-1}{10-1} = \frac{10}{9}(10^n-1)\). So \(S = \frac{5}{9}[\frac{10}{9}(10^n-1) - n] = \frac{5}{81}[10(10^n-1)-9n] = \frac{5}{81}(10^{n+1}-10-9n)\).

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82.

The number of terms in the A.P. 2, 5, 8, ..., 59 is:

A.19
B.20
C.21
D.22

Answer: B

\[ a_{n} = a + (n-1)d \quad\Longrightarrow\quad 59 = 2 + (n-1)3 \]

\[ 57 = 3(n-1) \quad\Longrightarrow\quad n = 20 \]

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83.

Insert 4 arithmetic means between 4 and 19.

A.7, 10, 13, 16
B.6, 9, 12, 15
C.5, 8, 11, 14
D.8, 11, 14, 17

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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84.

The sum of the series \(1 + 2x + 3x^2 + 4x^3 + ...\) for \(|x|<1\) is:

A.\(\frac{1}{(1-x)}\)
B.\(\frac{1}{(1+x)^2}\)
C.\(\frac{1}{(1-x)^2}\)
D.\(\frac{x}{(1-x)^2}\)

Answer: C

This is an Arithmetico-Geometric Progression (A.G.P.) with the A.P. part being 1, 2, 3, ... and G.P. part being \(1, x, x^2, ...\). The sum of an infinite A.G.P. is given by \(S = \frac{a}{1-r} + \frac{dr}{(1-r)^2}\). Here, the first term of the A.P. is \(a_{AP}=1\), \(d=1\), and \(r=x\). The first term of the A.G.P. is \(a=1\). The sum is \(S = \frac{1}{1-x} + \frac{1 \cdot x}{(1-x)^2} = \frac{1-x+x}{(1-x)^2} = \frac{1}{(1-x)^2}\). This is a standard result for the expansion of \((1-x)^{-2}\).

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85.

What term of the G.P. \(2, 1, 1/2, 1/4, ...\) is 1/128?

A.8th
B.9th
C.10th
D.11th

Answer: B

Here, \(a=2, r=1/2, a_n=1/128\). Using the formula \(a_n = ar^{n-1}\): \(1/128 = 2 \times (1/2)^{n-1} \Rightarrow 1/256 = (1/2)^{n-1}\). Since \(256 = 2^8\), we have \((1/2)^8 = (1/2)^{n-1}\). Equating powers: \(8=n-1 \Rightarrow n=9\). It is the 9th term.

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86.

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.

A.729
B.629
C.829
D.929

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

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

Important Instructions

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Other Subjects

Tags: A.P. G.P. H.P. Progressions Questions and Answers || Arithmetic Progression MCQ Questions || Geometric Progression GK Questions || Harmonic Progression MCQ || Quantitative Aptitude Progressions GK

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