JEE Mathematics Sequence and Series MCQs Set D

Question. If \( x^2 + 9y^2 + 25z^2 = xyz \left( \frac{15}{x} + \frac{5}{y} + \frac{3}{z} \right) \), then x, y and z are in
(a) AGP
(b) GP
(c) AP
(d) HP
Answer: (d) HP
Solution:
\( x^2 + 9y^2 + 25z^2 = xyz \left( \frac{15}{x} + \frac{5}{y} + \frac{3}{z} \right) \)
\( (x)^2 + (3y)^2 + (5z)^2 - 3xy - 15yz - 5zx = 0 \)
\( \Rightarrow \frac{1}{2} [(x - 3y)^2 + (3y - 5z)^2 + (5z - x)^2] = 0 \)
\( \Rightarrow x = 3y \) & \( 3y = 5z \) & \( 5z = x \)
\( \Rightarrow x = 3y = 5z \Rightarrow y = \frac{x}{3}, z = \frac{x}{5} \)
\( x, y, z \Rightarrow x, \frac{x}{3}, \frac{x}{5} \)
We know 1, 3, 5 in A.P. \( x \neq 0 \)
\( \Rightarrow \frac{1}{x}, \frac{3}{x}, \frac{5}{x} \) in A.P. \( \Rightarrow x, \frac{x}{3}, \frac{x}{5} \), in H.P.
\( \Rightarrow x, y, z \) in H.P.

Question. The sum to n term of the series
1(1!) + 2(2!) + 3(3!) + ....
(a) (n + 1)! - 1
(b) (n - 1)! - 1
(c) (n - 1)! + 1
(d) (n + 1)! + 1
Answer: (a) (n + 1)! - 1
Solution:
1(1!) + 2(2!) + 3(3!) + ... + n(n!)
\( \sum_{r=1}^n r(r!) = \sum_1^n (r + 1 - 1)r! = \sum_1^n [(r + 1)r! - r!] \)
\( = \sum_1^n [(r + 1)! - r!] = (n + 1)! - 1! = (n + 1)! - 1 \)

Question. The sum of all possible products of first n natural numbers taken two by two is
(a) \( \frac{1}{24} n(n+1)(n-1)(3n+2) \)
(b) \( \frac{n(n+1)(2n+1)}{6} \)
(c) \( \frac{n(n+1)(2n-1)(n+3)}{24} \)
(d) None of the options
Answer: (a) \( \frac{1}{24} n(n+1)(n-1)(3n+2) \)
Solution:
S = 1 . (2 + 3 + 4 + .... + n) +
+ 2 (3 + 4 + 5 + .... + n) +
+ 3 (4 + 5 + ..... + n) +
- - - - - - - - - - - - - - - - - -
+ (n - 2) [(n + 1) + n]
+ (n - 1) (n)
\( S = \frac{(1 + 2 + 3 + 4 + \dots + n)^2 - (1^2 + 2^2 + 3^2 + \dots + n^2)}{2} \)
\( = \frac{1}{2} \left[ \frac{n^2(n + 1)^2}{2^2} - \frac{n(n + 1)(2n + 1)}{6} \right] \)
\( = \frac{1}{24} n(n + 1) [3n^2 + 3n - 4n - 2] \)
\( = \frac{1}{24} n(n + 1)(3n^2 - n - 2) = \frac{1}{24} n(n + 1)(n - 1)(3n + 2) \)
Aliter
1 . 2 + 1.3 + 1.4 + ..... + 1. n
+ 2.3 + 2.4 + ...... + 2. n
+ 3.4 + ...... + 3. n
S = (1 . 2) + (1 + 2) 3 + (1 + 2 + 3) .4 + ... + [1 + 2 + 3 +.......+ (n - 1)]n
\( T_n = [1 + 2 + 3 + \dots (n - 1)] n \)
\( = \Sigma \frac{n^2(n - 1)}{2} = \frac{1}{2} (\Sigma n^3 - \Sigma n^2) \)

Question. The sum to 10 terms of the series
\( \sqrt{2} + \sqrt{6} + \sqrt{18} + \sqrt{54} + \dots \) is
(a) \( 121 (\sqrt{6} + \sqrt{2}) \)
(b) \( \frac{121}{2}(\sqrt{3} + 1) \)
(c) \( 243(\sqrt{3} + 1) \)
(d) \( 243(\sqrt{3} - 1) \)
Answer: (a) \( 121 (\sqrt{6} + \sqrt{2}) \)
Solution:
\( \sqrt{2} + \sqrt{6} + \sqrt{18} + \sqrt{54} + \dots \) + (10 terms)
\( = \sqrt{2} (1 + \sqrt{3} + \sqrt{9} + \sqrt{27} + \dots \) + (10 terms))
\( = \sqrt{2} (1 + 3^{1/2} + 3^1 + 3^{3/2} + \dots ) \)
\( = \sqrt{2} \cdot 1 \cdot \frac{(1 - (\sqrt{3})^{10})}{(1 - \sqrt{3})} = \frac{\sqrt{2}((\sqrt{3})^{10} - 1)}{(\sqrt{3} - 1)} \)
\( = \frac{\sqrt{2}(3^5 - 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{\sqrt{2}}{2} \cdot 242 (\sqrt{3} + 1) \)
\( = \sqrt{2} (121) (\sqrt{3} + 1) = 121 (\sqrt{6} + \sqrt{2}) \)

Question. If p is positive, then the sum to infinity of the series,
\( \frac{1}{1+p} - \frac{1-p}{(1+p)^2} + \frac{(1-p)^2}{(1+p)^3} - \dots \) is
(a) 1/2
(b) 3/4
(c) 1
(d) None of the options
Answer: (a) 1/2
Solution:
\( \frac{1}{(1+p)} - \frac{1-p}{(1+p)^2} + \frac{(1-p)^2}{(1+p)^3} - \dots \)
\( -1 < r = -\left( \frac{1-p}{1+p} \right) < 1 \quad \because p > 0 \)
\( = \frac{a}{1 - r} = \frac{\frac{1}{1+p}}{1 - \left(-\frac{1-p}{1+p}\right)} = \frac{1}{1+p+1-p} = \frac{1}{2} \)

Question. If \( G_1 \) and \( G_2 \) are two geometric means and A is the arithmetic means inserted between two positive numbers then the value of \( \frac{G_1^2}{G_2} + \frac{G_2^2}{G_1} \) is
(a) A/2
(b) A
(c) 2A
(d) None of the options
Answer: (c) 2A
Solution:
a, \( G_1, G_2 \), b & a, A, b
\( b = a r^3 \Rightarrow r = \left( \frac{b}{a} \right)^{1/3} \quad \because 2A = a + b \)
\( G_1 = ar = a \left( \frac{b}{a} \right)^{1/3} \quad \because G_1 G_2 = ab \)
\( G_2 = a r^2 = a \left( \frac{b}{a} \right)^{2/3} \)
\( \frac{G_1^2}{G_2} + \frac{G_2^2}{G_1} = \frac{G_1^3 + G_2^3}{G_1 G_2} = \frac{a^3 \frac{b}{a} + a^3 \frac{b^2}{a^2}}{ab} \)
\( = \frac{a^2 b + ab^2}{ab} = a + b = 2A \)

Question. \( \{a_n\} \) and \( \{b_n\} \) are two sequences given by
\( a_n = (x)^{1/2^n} + (y)^{1/2^n} \) and \( b_n = (x)^{1/2^n} - (y)^{1/2^n} \)
for all \( n \in N \). The value of \( a_1 a_2 a_3 \dots a_n \) is equal to
(a) x - y
(b) \( \frac{x + y}{b_n} \)
(c) \( \frac{x - y}{b_n} \)
(d) \( \frac{xy}{b_n} \)
Answer: (c) \( \frac{x - y}{b_n} \)
Solution:
\( a_n = x^{1/2^n} + y^{1/2^n} \) & \( b_n = x^{1/2^n} - y^{1/2^n} \), \( \forall n \in N \)
\( \Rightarrow a_n b_n = (x^{1/2^n})^2 - (y^{1/2^n})^2 \)
\( \Rightarrow a_n b_n = x^{1/2^{n-1}} - y^{1/2^{n-1}} = b_{n-1} \)
\( a_1 a_2 a_3 \dots a_n \times \frac{b_1 b_2 b_3 \dots b_n}{b_1 b_2 b_3 \dots b_n} \)
\( = \frac{(a_1 b_1)(a_2 b_2)(a_3 b_3)\dots(a_n b_n)}{b_1 b_2 b_3 \dots b_n} \)
\( = \frac{(x - y)(b_1)(b_2)(b_3)\dots(b_{n-1})}{b_1 b_2 b_3 \dots b_{n-1} \cdot b_n} = \frac{x - y}{b_n} \)

Question. The positive integer n for which
\( 2 \times 2^2 + 3 \times 2^3 + 4 \times 2^4 + \dots + n \times 2^n = 2^{n+10} \) is
(a) 510
(b) 511
(c) 512
(d) 513
Answer: (d) 513
Solution:
\( S = 2 \cdot 2^2 + 3 \cdot 2^3 + 4 \cdot 2^4 + \dots + n \cdot 2^n \)
\( 2S = 2 \cdot 2^3 + 3 \cdot 2^4 + \dots + (n - 1)2^n + n \cdot 2^{n+1} \)
\( \Rightarrow -S = 2^3 + 2^3 + 2^4 + \dots + 2^n - 2 \cdot n \cdot 2^{n+1} \)
\( \Rightarrow -S = 1 + \left( \frac{1(2^{n+1} - 1)}{2 - 1} \right) - n \cdot 2^{n+1} \)
\( -S = 1 + 2^{n+1} - 1 - n \cdot 2^{n+1} \)
\( S = n \cdot 2^{n+1} - 2^{n+1} = 2^{n+1}(n - 1) = 2^n \cdot 2^{10} \)
\( \Rightarrow n - 1 = 2^9 \Rightarrow n = 1 + 512 \Rightarrow n = 513 \)

Question. If \( 1^2 + 2^2 + 3^2 + \dots + 2003^2 = (2003)(4007)(334) \) and \( (1)(2003) + (2)(2002) + (3)(2001) + \dots + (2003)(1) = (2003)(334)(x). \), then x equals
(a) 2005
(b) 2004
(c) 2003
(d) 2001
Answer: (a) 2005
Solution:
\( 1^2 + 2^2 + 3^2 + \dots + 2003^2 = (2003) (4007) (334) \)
& \( 1 (2003) + 2 (2002) + 3 (2001) + \dots + 2003 (1) \)
= (2003) (334) (x)
\( \Rightarrow \sum_{r=1}^{2003} r[2003 - (r - 1)] = (2003) (334) x \)
\( \Rightarrow \sum r[2004 - r] = (2003) (334) x \)
\( \Rightarrow 2004 \sum_1^{2003} r - \sum_1^{2003} r^2 = (2003) (334) x \)
\( \Rightarrow \frac{(2003)(2004)^2}{2} - (2003) (4007) (334) = (2003) (334) x \)
\( \Rightarrow (2004) (1002) - (4007) (334) = (334)x \)
\( \Rightarrow 6(1002) - (4007) = x \)
\( \Rightarrow x = 6012 - 4007 = 2005 \)

Question. If x > 0, and \( \log_2 x + \log_2 (\sqrt{x}) + \log_2 (\sqrt[4]{x}) + \log_2 (\sqrt[8]{x}) + \log_2 (\sqrt[16]{x}) + \dots = 4 \), then x equals
(a) 2
(b) 3
(c) 4
(d) 5
Answer: (c) 4
Solution:
If x > 0
\( \log_2 x + \log_2 (\sqrt{x}) + \log_2 (\sqrt[4]{x}) + \log_2 (\sqrt[8]{x}) + \log_2 (\sqrt[16]{x}) + \dots = 4 \)
\( \Rightarrow \log_2 x + \log_2 x^{1/2} + \log_2 x^{1/4} + \dots = 4 \)
\( \Rightarrow (\log_2 x) \left[ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots \right] = 4 \)
\( \Rightarrow (\log_2 x) \frac{1}{1 - \frac{1}{2}} = 4 \Rightarrow \log_2 x = 2 \Rightarrow x = 4 \)

Question. If \( \sum_{r=1}^n t_r = \frac{1}{12} n(n+1)(n+2) \), the value \( \sum_{r=1}^n \frac{1}{t_r} \) is
(a) \( \frac{2n}{n+1} \)
(b) \( \frac{n}{(n+1)} \)
(c) \( \frac{4n}{n+1} \)
(d) \( \frac{3n}{n+1} \)
Answer: (c) \( \frac{4n}{n+1} \)
Solution:
\( \sum_{r=1}^n t_r = \frac{1}{12} n(n + 1) (n + 2) = S_n \)
\( t_r = S_n - S_{n-1} \)
\( = \frac{n(n + 1)(n + 2)}{12} - \frac{(n - 1)n(n + 1)}{12} \)
\( = \frac{n(n + 1)}{12} [n + 2 - n + 1] \)
\( t_r = \frac{n(n + 1)}{4} \Rightarrow \frac{1}{t_r} = \frac{4}{n(n + 1)} \)
\( \sum_{r=1}^n \frac{1}{t_r} = 4 \sum_{r=1}^n \frac{1}{n(n + 1)} = 4 \sum \left( \frac{1}{n} - \frac{1}{n + 1} \right) \)
\( = 4 \left[ 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \dots + \frac{1}{n} - \frac{1}{n + 1} \right] = \frac{4n}{n + 1} \)

Question. If a, b, c are in A.P. p, q, r are in H.P. and ap, bq, cr are in G.P., then \( \frac{p}{r} + \frac{r}{p} \) is equal to
(a) \( \frac{a}{c} + \frac{c}{a} \)
(b) \( \frac{a}{c} - \frac{c}{a} \)
(c) \( \frac{b}{q} + \frac{q}{b} \)
(d) \( \frac{b}{q} - \frac{a}{p} \)
Answer: (a) \( \frac{a}{c} + \frac{c}{a} \)
Solution:
a, b, c in A.P. \( \Rightarrow 2b = a + c \)
p, q, r, in H.P. \( \Rightarrow q = \frac{2pr}{p + r} \)
ap, bq, cr in G.P. \( \Rightarrow b^2 q^2 = acpr \)
\( \Rightarrow \frac{b^2 \cdot 4(pr)^2}{(p + r)^2} = ac \cdot pr \Rightarrow \frac{(a + c)^2 pr}{(p + r)^2} = ac \)
\( \Rightarrow \frac{(p + r)^2}{pr} = \frac{(a + c)^2}{ac} \Rightarrow \frac{p^2 + r^2}{pr} + 2 = \frac{a^2 + c^2}{ac} + 2 \)
\( \Rightarrow \frac{p}{r} + \frac{r}{p} = \frac{a}{c} + \frac{c}{a} \)

Question. The common difference d of the A.P. in which \( T_7 = 9 \) and \( T_1 T_2 T_7 \) is least is
(a) 33/2
(b) 5/4
(c) 33/20
(d) None of the options
Answer: (c) 33/20
Solution:
common diff. = d, in A.P.
\( T_7 = 9 \Rightarrow a + 6d = 9 \Rightarrow a = (9 - 6d) \)
\( T_1 T_2 T_7 = a \cdot (a + d) \cdot 9 = (9 - 6d) (9 - 5d) \cdot 9 \)
\( = 9 (30d^2 - 99d + 81) = 27 (10d^2 - 33d + 27) \)
Min value at \( d = \frac{-(-33)}{2 \cdot 10} = \frac{33}{20} \)

Question. The H.M. between two numbers is 16/5, their A.M. is A and G.M. is G. If \( 2A + G^2 = 26 \), then the numbers are
(a) 6, 8
(b) 4, 8
(c) 2, 8
(d) 1, 8
Answer: (c) 2, 8
Solution:
\( H = \frac{16}{5} = \frac{2ab}{a + b} \), \( A = \frac{a + b}{2} \), \( G = \sqrt{ab} \)
\( \frac{16}{5} = \frac{2G^2}{2A} \Rightarrow 5G^2 = 16A \)
& Given \( 2A + G^2 = 26 \)
\( \therefore 5(26 - 2A) = 16A \)
\( \Rightarrow 5 \cdot 26 = 26A \Rightarrow A = 5 \)
\( \therefore a + b = 10 \)
\( \frac{16}{5} = \frac{2ab}{10} \Rightarrow ab = 16 \Rightarrow a = 2, b = 8 \)

Question. \( 1^2 + 2^2 + \dots + n^2 = 1015 \), then value of n is
(a) 15
(b) 14
(c) 13
(d) None of the options
Answer: (b) 14
Solution:
\( 1^2 + 2^2 + \dots n^2 = 1015 \)
\( \frac{n(n + 1)(2n + 1)}{6} = 1015 \)
(A) \( n = 15 \Rightarrow \frac{15 \times 16 \times 31}{6} = 1240 \Rightarrow n \neq 15 \)
(B) \( n = 14 \Rightarrow \frac{14 \times 15 \times 29}{6} = 1015 \Rightarrow n = 14 \)

Question. If 1, 2, 3... are first terms; 1, 3, 5... are common differences and \( S_1, S_2, S_3 \dots \) are sums of n terms of given p AP's; then \( S_1 + S_2 + S_3 + \dots + S_p \) is equal to
(a) \( \frac{np(np + 1)}{2} \)
(b) \( \frac{n(np + 1)}{2} \)
(c) \( \frac{np(p + 1)}{2} \)
(d) \( \frac{np(np - 1)}{2} \)
Answer: (a) \( \frac{np(np + 1)}{2} \)
Solution:
\( S_1 = \frac{n}{2} [2 + (n - 1)1] \)
\( S_2 = \frac{n}{2} [4 + (n - 1)3] \)
\( S_3 = \frac{n}{2} [6 + (n - 1)5] \)
- - - - - - - - - - - - - - - - -
\( S_p = \frac{n}{2} [2p + (n - 1)(2p - 1)] \)
\( S_1 + S_2 + S_3 + \dots + S_p = \frac{n}{2} [(2 + 4 + 6 + \dots + 2p)] \)
\( + (n - 1) (1 + 3 + 5 + \dots + (2p - 1)] \)
\( = \frac{n}{2} [p (p + 1) + (n - 1) p^2] \)
\( = \frac{n}{2} [p^2 + p + np^2 - p^2] = \frac{np(1 + np)}{2} \)

Question. If a and b are \( p^{th} \) and \( q^{th} \) terms of an AP, then the sum of its (p + q) terms is
(a) \( \frac{p+q}{2} \left[a - b + \frac{a+b}{p-q}\right] \)
(b) \( \frac{p+q}{2} \left[a + b + \frac{a-b}{p-q}\right] \)
(c) \( \frac{p-q}{2} \left[a + b + \frac{a+b}{p+q}\right] \)
(d) None of the options
Answer: (b) \( \frac{p+q}{2} \left[a + b + \frac{a-b}{p-q}\right] \)
Solution:
\( T_p = a, T_q = b \)
\( a = A + (p - 1) d \)
\( b = A + (q - 1) d \Rightarrow \text{subtract } \frac{a - b}{p - q} = d \)
add \( a + b = 2A + (p + q - 1) d - d \)
\( \Rightarrow 2A + (p + q - 1) d = (a + b) + d \)
\( S_{p+q} = \frac{(p + q)}{2} [2A + (p + q - 1) d] \)
\( = \frac{(p + q)}{2} \left[ a + b + \frac{a - b}{p - q} \right] \)

Question. The sum of those integers from 1 to 100 which are not divisible by 3 or 5 is
(a) 2489
(b) 4735
(c) 2317
(d) 2632
Answer: (d) 2632
Solution:
S = 1 + 2 + 3 + ... + 100 = 5050
by 3, \( S_3 = \frac{33}{2} [3 + 99] = \frac{33 \times 102}{2} = 33 \times 51 = 1683 \)
by 5, \( S_5 = \frac{20}{2} [5 + 100] = 10 \times 105 = 1050 \)
by 3 & 5, \( S_{15} = \frac{6}{2} [15 + 90] = 3 \times 105 = 315 \)
by 3 or 5, \( S_{3 \text{ or } 5} = S_3 + S_5 - S_{15} \)
= 1683 + 1050 - 315 = 2418
Not divisible by 3 or 5 = 5050 - 2418 = 2632

Question. For the A.P. given by \(a_1, a_2, \ldots, a_n, \ldots\), the equations satisfied are
(a) \(a_1 + 2a_2 + a_3 = 0\)
(b) \(a_1 - 2a_2 + a_3 = 0\)
(c) \(a_1 + 3a_2 - 3a_3 - a_4 = 0\)
(d) \(a_1 - 4a_2 + 6a_3 - 4a_4 + a_5 = 0\)
Answer: (b) \(a_1 - 2a_2 + a_3 = 0\), (d) \(a_1 - 4a_2 + 6a_3 - 4a_4 + a_5 = 0\)

Solution:
\(a_1, a_2, \ldots, a_n, \ldots\)
\(a_2 = \frac{a_1 + a_3}{2} \Rightarrow a_1 + a_3 - 2a_2 = 0\)
\(a_1 - 2a_2 + a_3 = 0\)
\(-2(a_2 - 2a_3 + a_4) = 0\)
\(a_3 - 2a_4 + a_5 = 0\)
add \(a_1 - 4a_2 + 6a_3 - 4a_4 + a_5 = 0\)

Question. If sum of the infinite G.P., \(p, 1, \frac{1}{p}, \frac{1}{p^2}, \frac{1}{p^3}, \ldots\) is \(\frac{9}{2}\), the value of p is
(a) \(3\)
(b) \(\frac{3}{2}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{1}{3}\)
Answer: (a) \(3\), (c) \(\frac{2}{3}\)

Solution:
\(p + 1 + \frac{1}{p} + \frac{1}{p^2} + \frac{1}{p^3} + \ldots \infty = \frac{9}{2}\)
\(\Rightarrow \frac{p}{1 - \frac{1}{p}} = \frac{9}{2} \quad \because r = \frac{1}{p}, -1 < r < 1\)
\(\Rightarrow \frac{p^2}{p - 1} = \frac{9}{2}\)
\(\Rightarrow 2p^2 - 9p + 9 = 0\)
\(\Rightarrow (p - 3)(2p - 3) = 0\)
\(p = 3\), \(p = \frac{3}{2}\)

Question. If positive numbers a, b, c are in A.P. and \(a^2, b^2, c^2\) are in H.P., then
(a) \(a = b = c\)
(b) \(2b = a + c\)
(c) \(b^2 = \sqrt{\frac{ac}{8}}\)
(d) None of the options
Answer: (a) \(a = b = c\), (b) \(2b = a + c\)

Solution:
\(2b = a + c\)
\(4b^2 = a^2 + c^2 + 2ac\)
\(\Rightarrow a^2 + c^2 = 4b^2 - 2ac\)
& \(b^2 = \frac{2a^2c^2}{a^2 + c^2}\)
\(\Rightarrow b^2(4b^2 - 2ac) = 2a^2c^2\)
\(\Rightarrow b^2(2b^2 - ac) = a^2c^2\)
\(\Rightarrow 2b^4 - b^2(ac) - (ac)^2 = 0\)
\(\Rightarrow (b^2 - ac)(2b^2 + ac) = 0\)
\(\Rightarrow b^2 = ac\)
\(\Rightarrow a, b, c\) in G.P. & \(a, b, c\) in A.P.
\(\Rightarrow a = b = c\)
or \(2b^2 + ac = 0 \Rightarrow b^2 = -\frac{ac}{2} \Rightarrow a, b, \frac{-c}{2}\) in G.P.

Question. If the arithmetic mean of two positive numbers a & b (\(a > b\)) is twice their geometric mean, then \(a : b\) is
(a) \(2 + \sqrt{3} : 2 - \sqrt{3}\)
(b) \(7 + 4\sqrt{3} : 1\)
(c) \(1 : 7 - 4\sqrt{3}\)
(d) \(2 : \sqrt{3}\)
Answer: (a) \(2 + \sqrt{3} : 2 - \sqrt{3}\), (b) \(7 + 4\sqrt{3} : 1\), (c) \(1 : 7 - 4\sqrt{3}\)

Solution:
\(a, b > 0\), \(a > b\)
\(A = 2G\)
\(\frac{a + b}{2} = 2\sqrt{ab} \Rightarrow \frac{a + b}{\sqrt{ab}} = 4\)
\(\Rightarrow (a - b)^2 = 12ab \Rightarrow \frac{a - b}{\sqrt{ab}} = 2\sqrt{3}\)
\(\Rightarrow \frac{a + b}{a - b} = \frac{2}{\sqrt{3}}\)
C & D apply
\(\Rightarrow \frac{a + b + a - b}{a + b - a + b} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}}\)
\(\Rightarrow \frac{a}{b} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}}\)
Rationalize
\(\Rightarrow \frac{a}{b} = \frac{(2 + \sqrt{3})^2}{1} = \frac{7 + 4\sqrt{3}}{1} = \frac{1}{7 - 4\sqrt{3}}\)

Question. If \(\sum_{r=1}^{n} r(r+1)(2r+3) = an^4 + bn^3 + cn^2 + dn + e\), then
(a) \(a + c = b + d\)
(b) \(e = 0\)
(c) \(a, b - 2/3, c - 1\) are in A.P.
(d) \(c/a\) is an integer
Answer: (a) \(a + c = b + d\), (b) \(e = 0\), (c) \(a, b - 2/3, c - 1\) are in A.P., (d) \(c/a\) is an integer

Solution:
If \(\sum_{r=1}^{n} r(r+1)(2r+3) = an^4 + bn^3 + cn^2 + dn + e\)
\(= \sum (2r^3 + 5r^2 + 3r)\)
\(= \frac{2n^2(n+1)^2}{4} + \frac{5n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{2}\)
\(= \frac{n(n+1)}{2} \left[ n(n+1) + \frac{5}{3}(2n+1) + 3 \right]\)
\(= \frac{n(n+1)}{6} [3n^2 + 3n + 10n + 5 + 9]\)
\(= \frac{n(n+1)}{6} [3n^2 + 13n + 14] = \frac{1}{6}[3n^4 + 16n^3 + 27n^2 + 14n]\)
\(= \frac{1}{2}n^4 + \frac{8}{3}n^3 + \frac{9}{2}n^2 + \frac{7}{3}n + 0\)
Here, \(a = \frac{1}{2}, b = \frac{8}{3}, c = \frac{9}{2}, d = \frac{7}{3}, e = 0\)
(a) \(a + c = b + d\)
(b) \(e = 0\)
(d) \(\frac{c}{a} = \frac{9/2}{1/2} = 9 \in I\)
(c) \(a, b - \frac{2}{3}, c - 1 \Rightarrow \frac{1}{2}, 2, \frac{7}{2}\) in A.P.

Question. If \(b_1, b_2, b_3\) (\(b_i > 0\)) are three successive terms of a G.P. with common ratio r, the value of r for which the inequality \(b_3 > 4b_2 - 3b_1\), holds is given by
(a) \(r > 3\)
(b) \(0 < r < 1\)
(c) \(r = 3.5\)
(d) \(r = 5.2\)
Answer: (a) \(r > 3\), (b) \(0 < r < 1\), (c) \(r = 3.5\), (d) \(r = 5.2\)

Solution:
\(b_1, b_2, b_3 > 0\) in a G.P.
\(b_2^2 = b_1b_3\)
\(\Rightarrow \frac{b_2}{b_1} = \frac{b_3}{b_2} = r > 0\)
\(b_3 > 4b_2 - 3b_1 \quad \because \frac{b_1}{b_3} = \frac{1}{r^2}\)
\(\Rightarrow 1 > 4\frac{b_2}{b_3} - \frac{3b_1}{b_3} \Rightarrow 1 > \frac{4}{r} - \frac{3}{r^2} \quad \because r > 0\)
\(\Rightarrow r^2 - 4r + 3 > 0\)
\(\Rightarrow (r - 3)(r - 1) > 0\)
\(\Rightarrow r > 3\) or \(0 < r < 1\)
\(\Rightarrow 3.5, 5.2 > 3\)

Question. The value of \(\sum_{r=1}^{n} \frac{1}{\sqrt{a + rx} + \sqrt{a + (r - 1)x}}\) is
(a) \(\frac{n}{\sqrt{a} + \sqrt{a + nx}}\)
(b) \(\frac{n}{\sqrt{a} - \sqrt{a + nx}}\)
(c) \(\frac{\sqrt{a + nx} - \sqrt{a}}{x}\)
(d) \(\frac{\sqrt{a} + \sqrt{a + nx}}{x}\)
Answer: (a) \(\frac{n}{\sqrt{a} + \sqrt{a + nx}}\), (c) \(\frac{\sqrt{a + nx} - \sqrt{a}}{x}\)

Solution:
\(\sum_{r=1}^{n} \frac{1}{\sqrt{a + rx} + \sqrt{a + (r - 1)x}}\)
\(= \sum_{1}^{n} \frac{\sqrt{a + rx} - \sqrt{a + (r - 1)x}}{x}\)
\(= \frac{1}{x} \sum_{1}^{n} (\sqrt{a + rx} - \sqrt{a + (r - 1)x})\)
\(= \frac{1}{x} [\sqrt{a + nx} - \sqrt{a}]\) (Rationalize)
\(= \frac{1}{x} \frac{a + nx - a}{\sqrt{a + nx} + \sqrt{a}} = \frac{n}{\sqrt{a + nx} + \sqrt{a}}\)

Question. Let a, x, b be in A.P; a, y, b be in G.P. and a, z, b be in H.P. If \(x = y + 2\) and \(a = 5z\) then
(a) \(y^2 = xz\)
(b) \(x > y > z\)
(c) \(a = 9, b = 1\)
(d) \(a = 1/4, b = 9/4\)
Answer: (a) \(y^2 = xz\), (b) \(x > y > z\), (c) \(a = 9, b = 1\)

Solution:
A.P., \(2x = a + b\) & \(x = y + 2\), \(a = 5z\)
G.P., \(y^2 = ab\)
H.P., \(z = \frac{2ab}{a + b}\)
\(\Rightarrow z = \frac{2y^2}{2x} \Rightarrow y^2 = xz\)
A.M. > G.M. > H.M.
\(x > y > z\)
\(\frac{a}{5} = \frac{2ab}{a + b}\)
\(a^2 + ab = 10ab\)
\(\Rightarrow a(a - 9b) = 0\)
\(\Rightarrow a \neq 0\) or \(a = 9b\)
\(y = x - 2\)
\(\therefore y = \frac{2x - 4}{2} \Rightarrow y^2 = \frac{(a + b - 4)^2}{4}\)
\(\Rightarrow a^2 + b^2 - 2ab - 8a - 8b + 16 = 0\)
\(\Rightarrow 4b^2 - 5b + 1 = 0\)
\(\Rightarrow (b - 1)(4b - 1) = 0\)
\(\Rightarrow b = 1, a = 9\)
or \(b = 1/4, a = 9/4\)