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

Answer: A

The sum of n terms of an A.P. is always a quadratic expression of the form \(An^2+Bn\) with no constant term. Since the given sum \(S_n = 2n^2+3n\) is of this form, the series is an A.P. We can verify this by finding the nth term: \(a_n = S_n - S_{n-1} = (2n^2+3n) - [2(n-1)^2+3(n-1)] = (2n^2+3n) - [2(n^2-2n+1)+3n-3] = 4n+1\). The common difference is \(a_{n} - a_{n-1} = (4n+1) - (4(n-1)+1) = 4\), which is constant.

Answer: B

Given \(2b = a+c\), \(c^2=bd\), and \(c,d,e\) are in H.P. so \(d = \frac{2ce}{c+e}\). From \(c^2=bd\), we have \(d=c^2/b\). So \(c^2/b = \frac{2ce}{c+e} \Rightarrow c/b = \frac{2e}{c+e} \Rightarrow c(c+e) = 2be\). From the first condition, \(b=(a+c)/2\). Substitute this: \(c(c+e)=2(\frac{a+c}{2})e \Rightarrow c^2+ce = (a+c)e = ae+ce \Rightarrow c^2 = ae\). This is the condition for a, c, e to be in G.P.

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 \(S = \frac{5}{9}[\frac{10}{9}(10^n-1) - n] = \frac{5}{81}[10(10^n-1)-9n] = \frac{5}{81}(10^{n+1}-10-9n)\).

Answer: B

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

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

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: \(a+d=7\), \(a+2d=10\), \(a+3d=13\), \(a+4d=16\).

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 \((1-x)^{-2}\).

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.

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 \(S_{27} = \frac{27}{2}[2(1)+(26)2] = \frac{27}{2}[2+52] = \frac{27}{2}(54) = 27 \times 27 = 729\).

Answer: B

The sum is 7/2.

Apart from the leading 2, the terms are the powers of 1/2 and the powers of 1/3 interleaved. Summing each geometric series separately:

\[ \begin{aligned} \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} + \cdots &= \frac{1/2}{1 - 1/2} = 1 \\ \tfrac{1}{3} + \tfrac{1}{9} + \tfrac{1}{27} + \cdots &= \frac{1/3}{1 - 1/3} = \tfrac{1}{2} \\ 2 + 1 + \tfrac{1}{2} &= \tfrac{7}{2} \end{aligned} \]

Rearranging is allowed because every term is positive. 11/3, 13/4 and 9/2 come from mis-summing one of the two series.

Answer: A

The result is 1.

The sum of the first \(n\) odd numbers is \(n^{2}\): \(1 + 3 + 5 + \cdots + (2n - 1) = n^{2}\). For \(n = 20\) the sum is 400, and \(400/20^{2} = 1\).

Equivalently, the 20 odd numbers from 1 to 39 have average \((1 + 39)/2 = 20\), so their sum is \(20 \times 20\). The options 20, 2 and 1/2 come from dividing by 20 instead of \(20^{2}\) or from miscounting the terms.