Sequence and Series JEE Mathematics Worksheets Set 02

Question. Show that \( \ln (4 \times 12 \times 36 \times 108 \times \dots \text{ up to } n \text{ terms}) = 2n \ln 2 + \frac{n(n-1)}{2} \ln 3 \).
Answer: \( \ln (4 \times 12 \times 36 \times 108 \times \dots \text{ upto } n \text{ terms}) \)
\( = \ln \{(4) \times (4 \cdot 3) \times (4 \cdot 3^2) \times (4 \cdot 3^3) \times \dots \times (4 \times 3^{n-1})\} \)
\( = \ln (4^n \times 3^{1+2+3+\dots+(n-1)}) \)
\( = \ln 4^n + \ln(3^{\frac{n(n-1)}{2}}) \)
\( = 2n \ln 2 + \frac{n(n-1)}{2} \ln 3 \)

Question. There are \( n \) AM's between 1 & 31 such that 7th mean : \( (n - 1) \)th mean = 5 : 9, then find the value of \( n \).
Answer: 1, \( A_1, A_2, \dots, A_n \), 31 & \( \frac{A_7}{A_{n-1}} = \frac{5}{9} \)
\( d = \frac{31 - 1}{n + 1} = \frac{30}{n + 1} \)
\( \frac{A_7}{A_{n-1}} = \frac{1 + 7d}{1 + (n - 1)d} = \frac{1 + \frac{30}{n + 1} \times 7}{1 + \frac{(n - 1)30}{n + 1}} = \frac{5}{9} \)
\( \frac{n + 211}{31n - 29} = \frac{5}{9} \)
\( \implies 9n + 1899 = 155n - 145 \)
\( \implies 146n = 2044 \)
\( \implies n = 14 \)

Question. Prove that the average of the numbers \( n \sin n^\circ \), \( n = 2, 4, 6, \dots, 180 \) is \( \cot 1^\circ \).
Answer: \( \sum_{n=2,4,6}^{180} n \sin n^\circ = \frac{1}{90} [2 \sin 2^\circ + 4 \sin 4^\circ + 6 \sin 6^\circ + \dots + 178 \sin 178^\circ + 0] \)
\( = \frac{1}{90} [180 \sin 2^\circ + 180 \sin 4^\circ + \dots + 180 \sin 88^\circ + 90 \sin 90^\circ] \)
\( = 2 [\sin 2^\circ + \sin 4^\circ + \sin 6^\circ + \dots + \sin 88^\circ] + 1 \)
\( = 2 \left[ \frac{\sin 44^\circ \sin 45^\circ}{\sin 1^\circ} \right] + 1 \)
\( = \frac{\cos 1^\circ - \cos 89^\circ}{\sin 1^\circ} + 1 = \cot 1^\circ - 1 + 1 = \cot 1^\circ \)

Question. If S be the sum, P the product & R the sum of the reciprocals of n terms of a GP, find the value of \( P^2 \left( \frac{R}{S} \right)^n \).
Answer: \( S = a + ar + ar^2 + \dots + ar^{n-1} \implies S = \frac{a(1 - r^n)}{1 - r} \)
\( P = a^n r^{(1 + 2 + \dots + (n - 1))} \implies P = a^n r^{\frac{n(n-1)}{2}} \)
\( R = \frac{1}{a} + \frac{1}{ar} + \dots + \frac{1}{ar^{n-1}} = \frac{1}{ar^{n-1}}(1 + r + r^2 + \dots + r^{n-1}) \)
\( \implies R = \frac{1}{a^2 r^{n-1}} (a + ar + \dots + ar^{n-1}) \)
\( \implies R = \frac{1}{a^2 r^{n-1}} S \implies \frac{S}{R} = a^2 r^{n-1} \)
\( \implies \left( \frac{R}{S} \right)^n = \frac{1}{a^{2n} r^{n(n-1)}} \)
\( \implies \left( \frac{R}{S} \right)^n = \frac{1}{P^2} \)
\( \implies P^2 \left( \frac{R}{S} \right)^n = 1 \)

Question. In a set of four numbers, the first three are in GP & the last three are in AP, with common difference 6. If the first number is the same as the fourth, find the four numbers.
Answer: Let the last three terms be \( a - 6, a, a + 6 \).
The four numbers are \( (a + 6), (a - 6), a, (a + 6) \) since the first and fourth are equal.
First three terms are in GP: \( (a - 6)^2 = a(a + 6) \)
\( a^2 - 12a + 36 = a^2 + 6a \)
\( \implies 18a = 36 \)
\( \implies a = 2 \)
Set of numbers are {8, -4, 2, 8}.

Question. Find the sum of the series, \( 7+77+777 + \dots \) to n terms.
Answer: \( S_n = 7 + 77 + 777 + \dots + n \text{ terms} \)
\( S_n = \frac{7}{9} [9 + 99 + 999 + \dots + n \text{ terms}] \)
\( S_n = \frac{7}{9} [(10-1) + (10^2-1) + (10^3-1) + \dots + (10^n-1)] \)
\( = \frac{7}{9} [ (10 + 10^2 + \dots + 10^n) - n ] \)
\( = \frac{7}{9} \left[ \frac{10(10^n - 1)}{9} - n \right] \)
\( = \frac{7}{81} [10^{n+1} - 9n - 10] \)

Question. Find three numbers a, b, c between 2 & 18 such that ;
(i) their sum is 25
(ii) the numbers 2, a, b are consecutive terms of an AP &
(iii) the numbers b, c, 18 are consecutive terms of a GP.
Answer: 2, a, b, c, 18
\( a + b + c = 25 \) ......(i)
\( 2a = 2 + b \implies b = 2a - 2 \) ......(ii)
\( c^2 = 18b \) ......(iii)
From (i), \( a + (2a - 2) + c = 25 \implies c = 27 - 3a \)
Substitute \( b \) and \( c \) in (iii):
\( (27 - 3a)^2 = 18(2a - 2) \)
\( 9(9 - a)^2 = 36(a - 1) \)
\( (9 - a)^2 = 4(a - 1) \)
\( a^2 - 18a + 81 = 4a - 4 \)
\( \implies a^2 - 22a + 85 = 0 \)
\( \implies (a - 17)(a - 5) = 0 \)
\( \implies a = 17 \) or \( a = 5 \)
If \( a = 17 \), \( b = 32 \) (Reject as \( b < 18 \)).
If \( a = 5 \), \( b = 2(5) - 2 = 8 \).
\( c^2 = 18(8) = 144 \implies c = 12 \).
The numbers are 5, 8, 12.

Question. If one AM 'a' & two GM's p & q be inserted between any two given numbers then show that \( p^3 + q^3 = 2apq \).
Answer: Let the numbers be \( A \) and \( B \).
AM: \( a = \frac{A + B}{2} \implies 2a = A + B \)
GM's: \( A, p, q, B \) are in GP.
\( p = Ar, q = Ar^2, B = Ar^3 \)
Also \( p^2 = Aq \) and \( q^2 = pB \).
\( 2a = \frac{p^2}{q} + \frac{q^2}{p} \)
\( \implies 2a = \frac{p^3 + q^3}{pq} \)
\( \implies p^3 + q^3 = 2apq \)

Question. The \( 1^{st}, 2^{nd} \text{ and } 3^{rd} \) terms of an arithmetic series are a, b and \( a^2 \) where 'a' is negative. The \( 1^{st}, 2^{nd} \text{ and } 3^{rd} \) terms of a geometric series are a, \( a^2 \) and b find the
(i) value of a and b
(ii) sum of infinite geometric series if it exists. If no then find the sum to n terms of the GP.
(iii) sum of the 40 term of the arithmetic series.
Answer: (i) \( 2b = a + a^2 \) (from AP) and \( (a^2)^2 = ab \implies a^4 = ab \)
Since \( a \neq 0 \), \( a^3 = b \).
Substitute \( b = a^3 \) in \( 2b = a + a^2 \):
\( 2a^3 = a + a^2 \implies 2a^2 - a - 1 = 0 \)
\( \implies (a - 1)(2a + 1) = 0 \)
Since \( a \) is negative, \( a = -1/2 \).
Then \( b = (-1/2)^3 = -1/8 \).
(ii) GP: \( a, a^2, b \implies -1/2, 1/4, -1/8 \)
\( r = \frac{1/4}{-1/2} = -1/2 \).
\( S_{\infty} = \frac{a}{1 - r} = \frac{-1/2}{1 - (-1/2)} = \frac{-1/2}{3/2} = -1/3 \).
(iii) AP: \( a, b, a^2 \implies -1/2, -1/8, 1/4 \)
\( d = -1/8 - (-1/2) = 3/8 \).
\( S_{40} = \frac{40}{2} [2(-1/2) + 39(3/8)] \)
\( = 20 [-1 + 117/8] = 20 [109/8] = 545/2 \).

Question. If the 10th term of an HP is 21 & 21st term of the same HP is 10, then find the 210th term.
Answer: For the corresponding AP:
\( a + 9d = 1/21 \)
\( a + 20d = 1/10 \)
Subtracting: \( 11d = 1/10 - 1/21 = 11/210 \implies d = 1/210 \).
\( a = 1/21 - 9/210 = 10/210 - 9/210 = 1/210 \).
210th term of AP: \( a + 209d = 1/210 + 209/210 = 210/210 = 1 \).
So, 210th term of HP = 1.

Question. Find the sum of the n terms and to infinity of the sequence \( \frac{1}{1^2+1^4} + \frac{2}{1+2^2+2^4} + \frac{3}{1+3^2+3^4} + \dots \).
Answer: \( T_n = \frac{n}{1 + n^2 + n^4} = \frac{n}{(n^2 + n + 1)(n^2 - n + 1)} \)
\( T_n = \frac{1}{2} \left[ \frac{1}{n^2 - n + 1} - \frac{1}{n^2 + n + 1} \right] \)
\( S_n = \sum T_n = \frac{1}{2} \left[ (1 - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{7}) + \dots + (\frac{1}{n^2 - n + 1} - \frac{1}{n^2 + n + 1}) \right] \)
\( S_n = \frac{1}{2} \left[ 1 - \frac{1}{n^2 + n + 1} \right] = \frac{n^2 + n}{2(n^2 + n + 1)} \)
\( S_{\infty} = 1/2 \).

Question. An AP & an HP have the same first term, the same last term & the same number of terms; prove that the product of the \( r^{th} \) term from the beginning in one series & the \( r^{th} \) term from the end in the other is independent of r.
Answer: Let AP be \( a, a+d, \dots, a+(n-1)d \).
Let HP be \( \frac{1}{A}, \frac{1}{A+D}, \dots, \frac{1}{A+(n-1)D} \).
Given \( a = 1/A \) and \( a+(n-1)d = \frac{1}{A+(n-1)D} \).
\( r^{th} \) term of AP from beginning \( T_r = a + (r - 1)d \).
\( r^{th} \) term of HP from end is the \( (n - r + 1)^{th} \) term from beginning.
\( H'_r = \frac{1}{A + (n - r)D} \).
Product \( T_r \times H'_r = \frac{a + (r - 1)d}{A + (n - r)D} \).
Taking specific cases \( a=1, l=1 \), product is -1, which is independent of r.

Question. Find the sum of the first n terms of the sequence: \( 1 + 2(1 + \frac{1}{n}) + 3(1 + \frac{1}{n})^2 + 4(1 + \frac{1}{n})^3 + \dots \).
Answer: Let \( r = 1 + 1/n \).
\( S_n = 1 + 2r + 3r^2 + \dots + nr^{n-1} \)
\( r S_n = r + 2r^2 + \dots + (n-1)r^{n-1} + nr^n \)
Subtracting:
\( S_n(1 - r) = 1 + r + r^2 + \dots + r^{n-1} - nr^n \)
\( S_n(1 - r) = \frac{r^n - 1}{r - 1} - nr^n \)
Since \( 1 - r = -1/n \):
\( S_n(-1/n) = \frac{(1 + 1/n)^n - 1}{1/n} - n(1 + 1/n)^n \)
\( \implies S_n = n^2 \)

Question. Find the nth term and the sum of n terms of the sequence
(i) 1 + 5 + 13 + 29 + 61 + .....
(ii) 6 + 13 + 22 + 33 + ......
Answer: (i) \( T_n = 1 + 4(2^{n-1} - 1) = 2^{n+1} - 3 \)
\( S_n = \sum (2^{n+1} - 3) = 2(2^n - 1) - 3n = 2^{n+2} - 3n - 4 \).
(ii) Differences are 7, 9, 11, ... which is an AP.
\( T_n = 6 + [7 + 9 + \dots + (2n + 3)] = 6 + (n-1)(n+5) = n^2 + 4n + 1 \).
\( S_n = \sum (n^2 + 4n + 1) = \frac{n(n+1)(2n+1)}{6} + \frac{4n(n+1)}{2} + n = \frac{n(n+1)(2n+13)}{6} + n \).

Question. The AM of two numbers exceeds their GM by 15 & HM by 27. Find the numbers.
Answer: \( A - G = 15 \) and \( A - H = 27 \).
We know \( G^2 = AH \).
\( (A - 15)^2 = A(A - 27) \)
\( A^2 - 30A + 225 = A^2 - 27A \)
\( \implies 3A = 225 \implies A = 75 \).
Then \( G = 60 \).
\( a + b = 150 \), \( ab = 3600 \).
Solving \( x^2 - 150x + 3600 = 0 \):
\( (x - 120)(x - 30) = 0 \).
Numbers are 120 and 30.

Question. Sum the following series to n terms and to infinity:
(i) \( \frac{1}{1.4.7} + \frac{1}{4.7.10} + \frac{1}{7.10.13} + \dots \)
(ii) \( \sum_{r=1}^{n} r(r+1)(r+2)(r+3) \)
(iii) \( \sum_{r=1}^{n} \frac{1}{4r^2 - 1} \)
(iv) \( \frac{1}{4} + \frac{1.3}{4.6} + \frac{1.3.5}{4.6.8} + \dots \)
Answer: (i) \( S_n = \frac{1}{24} \left[ 1 - \frac{6}{(3n+1)(3n+4)} \right], S_{\infty} = \frac{1}{24} \).
(ii) \( S_n = \frac{n(n+1)(n+2)(n+3)(n+4)}{5} \).
(iii) \( T_r = \frac{1}{(2r-1)(2r+1)} = \frac{1}{2} \left[ \frac{1}{2r-1} - \frac{1}{2r+1} \right] \).
\( S_n = \frac{1}{2} \left[ 1 - \frac{1}{2n+1} \right] = \frac{n}{2n+1}, S_{\infty} = 1/2 \).
(iv) \( S_{\infty} = 1 \).

Question. Evaluate the sum \( \sum_{n=1}^{\infty} \frac{n^2}{6^n} \).
Answer: Let \( S = \frac{1^2}{6} + \frac{2^2}{6^2} + \frac{3^2}{6^3} + \dots \)
\( \frac{1}{6} S = \frac{1^2}{6^2} + \frac{2^2}{6^3} + \dots \)
Subtracting:
\( \frac{5}{6} S = \frac{1}{6} + \frac{3}{6^2} + \frac{5}{6^3} + \dots \)
This is an AGP. Applying the formula for sum of infinite AGP:
\( \frac{5}{6} S = \frac{1/6}{1 - 1/6} + \frac{2/6^2}{(1 - 1/6)^2} = \frac{1}{5} + \frac{2}{25} = \frac{7}{25} \)
\( S = \frac{7}{25} \times \frac{6}{5} = \frac{42}{125} \).

Question. If the sum \( \sqrt{1 + \frac{1}{1^2} + \frac{1}{2^2}} + \sqrt{1 + \frac{1}{2^2} + \frac{1}{3^2}} + \dots + \sqrt{1 + \frac{1}{1999^2} + \frac{1}{2000^2}} \) equal to \( n - 1/n \) where \( n \in N \). Find n.
Answer: General term \( T_r = \sqrt{1 + \frac{1}{r^2} + \frac{1}{(r+1)^2}} = \sqrt{\frac{r^2(r+1)^2 + (r+1)^2 + r^2}{r^2(r+1)^2}} \)
\( = \sqrt{\frac{r^2(r+1)^2 + 2r^2 + 2r + 1}{r^2(r+1)^2}} = \sqrt{\frac{(r^2 + r + 1)^2}{r^2(r+1)^2}} = \frac{r^2+r+1}{r(r+1)} = 1 + \frac{1}{r(r+1)} \)
\( T_r = 1 + \frac{1}{r} - \frac{1}{r+1} \)
Sum \( = \sum_{r=1}^{1999} (1 + \frac{1}{r} - \frac{1}{r+1}) = 1999 + (1 - \frac{1}{2000}) = 2000 - \frac{1}{2000} \).
Comparing with \( n - 1/n \), \( n = 2000 \).

Question. Show that in any arithmetic progression \( a_1, a_2, a_3, \dots \)
\( a_1^2 - a_2^2 + a_3^2 - a_4^2 + \dots + a_{2K-1}^2 - a_{2K}^2 = \frac{K}{2K-1} (a_1^2 - a_{2K}^2) \).
Answer: \( LHS = (a_1-a_2)(a_1+a_2) + (a_3-a_4)(a_3+a_4) + \dots \)
\( = -d(a_1+a_2) - d(a_3+a_4) - \dots - d(a_{2K-1}+a_{2K}) \)
\( = -d [a_1 + a_2 + \dots + a_{2K}] \)
\( = -d \frac{2K}{2} (a_1 + a_{2K}) = -Kd(a_1 + a_{2K}) \)
Since \( a_{2K} = a_1 + (2K-1)d \implies d = \frac{a_{2K}-a_1}{2K-1} \).
\( LHS = -K \frac{a_{2K}-a_1}{2K-1} (a_1 + a_{2K}) = \frac{K}{2K-1} (a_1^2 - a_{2K}^2) \).

Question. If the first 3 consecutive terms of a geometrical progression are the roots of the equation \( 2x^3 - 19x^2 + 57x - 54 = 0 \) find the sum to infinite number of terms of G.P.
Answer: Let the roots be \( a/r, a, ar \).
Product of roots: \( (a/r)(a)(ar) = 54/2 = 27 \implies a^3 = 27 \implies a = 3 \).
Sum of roots: \( 3/r + 3 + 3r = 19/2 \)
\( \implies 6 + 6r + 6r^2 = 19r \)
\( \implies 6r^2 - 13r + 6 = 0 \)
\( \implies (2r-3)(3r-2) = 0 \implies r = 3/2 \) or \( r = 2/3 \).
For sum to infinity, \( |r| < 1 \), so \( r = 2/3 \).
First term \( a' = a/r = 3 / (2/3) = 9/2 \).
\( S_{\infty} = \frac{9/2}{1 - 2/3} = \frac{9/2}{1/3} = 27/2 \).

Question. If the roots of \( 10x^3 - cx^2 - 54x - 27 = 0 \) are in harmonic progression, then find c & all the roots.
Answer: Let roots be \( \alpha, \beta, \gamma \) in HP. Then \( \beta = \frac{2\alpha\gamma}{\alpha+\gamma} \implies \alpha\beta + \beta\gamma + \gamma\alpha = 3\alpha\gamma \).
From equation: \( \sum \alpha\beta = -54/10 = -27/5 \).
\( 3\alpha\gamma = -27/5 \implies \alpha\gamma = -9/5 \).
Product \( \alpha\beta\gamma = 27/10 \implies (-9/5)\beta = 27/10 \implies \beta = -3/2 \).
Sum \( \alpha+\beta+\gamma = c/10 \).
Since \( \beta = -3/2 \) is a root, \( 10(-3/2)^3 - c(-3/2)^2 - 54(-3/2) - 27 = 0 \)
\( -135/4 - 9c/4 + 81 - 27 = 0 \implies c = 9 \).
Roots are 3, -3/2, -3/5.

Question. If a, b, c, d, e be 5 numbers such that a, b, c are in AP; b, c, d are in GP & c, d, e are in HP then
(i) Prove that a, c, e are in GP.
(ii) Prove that \( e = (2b - a)^2/a \).
(iii) If a = 2 & e = 18, find all possible values of b, c, d.
Answer: (i) \( 2b = a+c \), \( c^2 = bd \), \( d = \frac{2ce}{c+e} \).
\( c^2 = b \left( \frac{2ce}{c+e} \right) \implies c = \frac{2be}{c+e} \implies c^2 + ce = 2be \).
Substituting \( 2b = a+c \): \( c^2 + ce = (a+c)e = ae + ce \implies c^2 = ae \).
Hence \( a, c, e \) are in GP.
(ii) \( c = 2b - a \). Since \( c^2 = ae \), \( (2b - a)^2 = ae \implies e = (2b - a)^2/a \).
(iii) \( a=2, e=18 \implies c^2 = 2 \times 18 = 36 \implies c = 6 \) or \( -6 \).
If \( c=6 \), \( 2b = 2+6=8 \implies b=4 \). \( d = 6^2/4 = 9 \).
If \( c=-6 \), \( 2b = 2-6=-4 \implies b=-2 \). \( d = (-6)^2/(-2) = -18 \).

Question. If n is a root of the equation \( x^2(1 - ac) - x(a^2 + c^2) - (1 + ac) = 0 \) & if n HM's are inserted between a & c, show that the difference between the first & the last mean is equal to ac (a - c).
Answer: The equation is \( (n^2-1) = (na+c)(nc+a) \).
For HM's \( H_1, \dots, H_n \), the common difference of the corresponding AP is \( d = \frac{a-c}{ac(n+1)} \).
\( \frac{1}{H_1} = \frac{1}{a} + d \implies H_1 = \frac{ac(n+1)}{cn+a} \).
\( \frac{1}{H_n} = \frac{1}{a} + nd \implies H_n = \frac{ac(n+1)}{an+c} \).
\( H_1 - H_n = ac(n+1) \left[ \frac{1}{nc+a} - \frac{1}{na+c} \right] \)
\( = \frac{ac(n+1)(n-1)(a-c)}{(nc+a)(na+c)} \).
Using the equation \( (nc+a)(na+c) = n^2-1 \):
\( H_1 - H_n = ac(a-c) \).

Question. (i) The value of x + y + z is 15 if a, x, y, z, b are in AP while the value of (1/x) + (1/y) + (1/z) is 5/3 if a, x, y, z, b are in HP. Find a & b.
(ii) The values of xyz is 15/2 or 18/5 according as the series a, x, y, z, b is an AP or HP. Find the values of a & b assuming them to be positive integer.
Answer: (i) In AP, \( x+y+z = 3(\frac{a+b}{2}) = 15 \implies a+b = 10 \).
In HP, \( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} = 3(\frac{1/a+1/b}{2}) = \frac{3(a+b)}{2ab} = 5/3 \).
\( \frac{3(10)}{2ab} = 5/3 \implies ab = 9 \).
Solving \( a+b=10, ab=9 \), we get \( \{a, b\} = \{1, 9\} \).
(ii) For AP case: \( xyz = \frac{15}{2} \). Using \( d = \frac{b-a}{4} \):
\( (a+d)(a+2d)(a+3d) = 15/2 \).
Substituting \( b=3, a=1 \) or similar integers. For \( a=1, b=3 \), \( d=1/2 \):
\( (1.5)(2)(2.5) = 7.5 = 15/2 \).
So \( a=1, b=3 \) or \( a=3, b=1 \).

Question. Prove that the sum of the infinite series \( \frac{1.3}{2} + \frac{3.5}{2^2} + \frac{5.7}{2^3} + \frac{7.9}{2^4} + \dots \infty = 23 \).
Answer: \( S = \frac{3}{2} + \frac{15}{4} + \frac{35}{8} + \frac{63}{16} + \dots \)
\( \frac{S}{2} = \frac{3}{4} + \frac{15}{8} + \frac{35}{16} + \dots \)
Subtracting: \( \frac{S}{2} = \frac{3}{2} + \frac{12}{4} + \frac{20}{8} + \frac{28}{16} + \dots \)
\( \frac{S}{2} = \frac{3}{2} + 3 + \frac{5}{2} + \frac{7}{4} + \dots \)
\( \frac{S}{2} = \frac{3}{2} + [3 + \frac{5}{2} + \frac{7}{4} + \dots] \)
The bracket is an AGP: \( S' = 3 + \frac{5}{2} + \frac{7}{4} + \dots \)
\( \frac{S'}{2} = \frac{3}{2} + \frac{5}{4} + \dots \)
\( \frac{S'}{2} = 3 + \frac{2}{2} + \frac{2}{4} + \dots = 3 + \frac{1}{1-1/2} = 5 \implies S' = 10 \).
\( \frac{S}{2} = 1.5 + 10 = 11.5 \implies S = 23 \).

Question. Find the condition that the roots of the equation \( x^3 + px^2 + qx - r = 0 \) may be in A.P. and hence solve the equation \( x^3 - 12x^2 + 39x - 28 = 0 \).
Answer: Let roots be \( a-d, a, a+d \).
Sum \( = 3a = -p \implies a = -p/3 \).
Since \( a \) is a root: \( (-p/3)^3 + p(-p/3)^2 + q(-p/3) - r = 0 \)
\( \implies 2p^3 - 9pq + 27r = 0 \).
For \( x^3 - 12x^2 + 39x - 28 = 0 \):
\( 3a = 12 \implies a = 4 \).
Product \( a(a^2-d^2) = 28 \implies 4(16-d^2) = 28 \implies 16-d^2 = 7 \implies d = \pm 3 \).
Roots are 1, 4, 7.

Question. If a, b, c be in GP & \( \log_c a, \log_b c, \log_a b \) be in AP, then show that the common difference of the AP must be 3/2.
Answer: \( b^2 = ac \). Let \( \log_b c = t \). Then \( \log_c a = \log_c (b^2/c) = 2\log_c b - 1 = \frac{2}{t} - 1 \).
\( \log_a b = \frac{1}{\log_b a} = \frac{1}{\log_b (b^2/c)} = \frac{1}{2-t} \).
AP condition: \( (\frac{2}{t}-1) + \frac{1}{2-t} = 2t \)
\( \implies \frac{2-t}{t} + \frac{1}{2-t} = 2t \)
\( \implies \frac{(2-t)^2 + t}{t(2-t)} = 2t \implies \frac{t^2-3t+4}{2t-t^2} = 2t \)
Solving for \( t \), we find \( d = t - (\frac{2}{t}-1) = 3/2 \).

Question. Two distinct, real infinite geometric series each have a sum of 1 and have the same second term. The third term of one of the series is 1/8. If the second term of both the series can be written in the form \( \frac{\sqrt{m} - n}{p} \), where m, n and p are positive integers and m is not divisible by the square of any prime, find the value of 100m + 10n + p.
Answer: Let series be \( a, ar, ar^2 \dots \) and \( A, AR, AR^2 \dots \).
\( \frac{a}{1-r} = 1 \), \( \frac{A}{1-R} = 1 \), \( ar = AR \), \( ar^2 = 1/8 \).
\( a = 1-r, A = 1-R \). \( (1-r)r = (1-R)R \implies r-r^2 = R-R^2 \implies R = 1-r \).
\( (1-r)r^2 = 1/8 \implies 8r^3 - 8r^2 + 1 = 0 \).
\( (2r-1)(4r^2-2r-1) = 0 \). Since series are distinct, \( r \neq 1/2 \).
\( r = \frac{1+\sqrt{5}}{4} \).
Second term \( ar = (1-r)r = \frac{\sqrt{5}-1}{8} \).
\( m=5, n=1, p=8 \).
Value \( = 100(5) + 10(1) + 8 = 518 \).

Question. One of the roots of the equation \( 2000x^6 + 100x^5 + 10x^3 + x - 2 = 0 \) is of the form \( \frac{\sqrt{m} - n}{r} \), where m is non zero integer and n and r are relatively prime natural numbers. Find the value of m + n + r.
Answer: \( 2000x^6 + 100x^5 + 10x^3 + x - 2 = 0 \)
\( \implies 2(1000x^6 - 1) + x \frac{1000x^6-1}{10x^2-1} = 0 \)
\( \implies (1000x^6 - 1) [2 + \frac{x}{10x^2-1}] = 0 \).
\( 20x^2 + x - 2 = 0 \implies x = \frac{-1 \pm \sqrt{1 + 160}}{40} = \frac{\sqrt{161}-1}{40} \).
\( m=161, n=1, r=40 \).
\( m+n+r = 161+1+40 = 202 \).

Question. In a GP the ratio of the sum of the first eleven terms to the sum of the last eleven terms is 1/8 and the ratio of the sum of all the terms without the first nine to the sum of all the terms without the last nine is 2. Find the number of terms in the G.P.
Answer: Let G.P. be \( a, ar, \dots, ar^{n-1} \).
\( \frac{a(r^{11}-1)/(r-1)}{ar^{n-11}(r^{11}-1)/(r-1)} = \frac{1}{8} \implies r^{n-11} = 8 \).
Also, \( \frac{S - S_9}{S - (\text{last } 9)} = 2 \implies r^9 = 2 \).
\( (2^{1/9})^{n-11} = 2^3 \)
\( \implies \frac{n-11}{9} = 3 \)
\( \implies n - 11 = 27 \implies n = 38 \).

Question. Given a three digit number whose digits are three successive terms of a G.P. If we subtract 792 from it, we get a number written by the same digits in the reverse order. Now if we subtract four from the hundred’s digit of the initial number and leave the other digits unchanged, we get a number whose digits are successive terms of an A.P. Find the number.
Answer: Let digits be \( x, y, z \). Number \( = 100x + 10y + z \). Digits in GP: \( y^2 = xz \).
\( (100x + 10y + z) - 792 = 100z + 10y + x \)
\( \implies 99(x - z) = 792 \implies x - z = 8 \).
Since digits, \( x=9, z=1 \).
Then \( y^2 = 9 \times 1 \implies y = 3 \).
Original number is 931.
Check AP: \( x-4, y, z \implies 5, 3, 1 \), which is in AP. Correct.