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Relation between A.P., G.P. Questions in English

Class 11 Mathematics · Sequences and Series · Relation between A.P., G.P.

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101
AdvancedMCQ
If $a, b, c \in \mathbb{R}^+$ are such that $2a, b, 4c$ are in $A.P.$ and $c, a, b$ are in $G.P.$,then:
A
$a^2, ac, c^2$ are in $A.P.$
B
$c, a, a+2c$ are in $A.P.$
C
$c, a, a+2c$ are in $G.P.$
D
$\frac{a}{2}, c, c-a$ are in $G.P.$

Solution

(C) Given $2a, b, 4c$ are in $A.P.$,we have $2b = 2a + 4c$,which simplifies to $b = a + 2c$.
Given $c, a, b$ are in $G.P.$,we have $a^2 = bc$.
Substituting $b = a + 2c$ into $a^2 = bc$,we get $a^2 = c(a + 2c) = ac + 2c^2$.
Rearranging gives $a^2 - ac - 2c^2 = 0$.
Factoring the quadratic: $(a - 2c)(a + c) = 0$.
Since $a, c \in \mathbb{R}^+$,we must have $a = 2c$.
Now,check the options:
For option $B$: The terms are $c, a, a+2c$. Substituting $a=2c$,we get $c, 2c, 2c+2c = c, 2c, 4c$.
These are in $G.P.$ with common ratio $2$,but let's check if they are in $A.P.$: $2c - c = c$ and $4c - 2c = 2c$. Since $c \neq 2c$,they are not in $A.P.$
For option $C$: The terms are $c, a, a+2c$. Substituting $a=2c$,we get $c, 2c, 4c$. These are in $G.P.$ with common ratio $2$.
102
AdvancedMCQ
The number of terms common between the two series $2 + 5 + 8 + \dots$ up to $50$ terms and the series $3 + 5 + 7 + 9 + \dots$ up to $60$ terms is:
A
$18$
B
$20$
C
$22$
D
$24$

Solution

(B) The first series is an Arithmetic Progression $(AP)$ with $a_1 = 2$,$d_1 = 3$,and $n_1 = 50$. The $50^{th}$ term is $T_{50} = 2 + (50 - 1)3 = 2 + 147 = 149$.
The second series is an $AP$ with $a_2 = 3$,$d_2 = 2$,and $n_2 = 60$. The $60^{th}$ term is $T'_{60} = 3 + (60 - 1)2 = 3 + 118 = 121$.
Common terms must satisfy both sequences. The first few common terms are $5, 11, 17, \dots$. This forms a new $AP$ with first term $A = 5$ and common difference $D = \text{lcm}(3, 2) = 6$.
The general term of the common series is $T_n = 5 + (n - 1)6 = 6n - 1$.
Since the common terms must be present in both series,$T_n \leq 121$ (the smaller of the two last terms).
$6n - 1 \leq 121 \implies 6n \leq 122 \implies n \leq 20.33$.
Since $n$ must be an integer,the number of common terms is $20$.
103
DifficultMCQ
If $n$ arithmetic means $a_1, a_2, \dots, a_n$ are inserted between $50$ and $100$ and $n$ harmonic means $h_1, h_2, \dots, h_n$ are inserted between the same two numbers,then $a_2 h_{n-1}$ is equal to
A
$5000$
B
$\frac{10000}{n}$
C
$10000$
D
$\frac{250}{n}$

Solution

(A) Let the numbers be $A = 50$ and $B = 100$.
For arithmetic means,the common difference $d = \frac{B-A}{n+1} = \frac{50}{n+1}$.
Then $a_2 = A + 2d = 50 + \frac{100}{n+1} = \frac{50n + 50 + 100}{n+1} = \frac{50(n+3)}{n+1}$.
For harmonic means,$\frac{1}{h_1}, \frac{1}{h_2}, \dots, \frac{1}{h_n}$ are in $A$.$P$. with first term $\frac{1}{A} = \frac{1}{50}$ and last term $\frac{1}{B} = \frac{1}{100}$.
The common difference $d' = \frac{\frac{1}{100} - \frac{1}{50}}{n+1} = \frac{-1}{100(n+1)}$.
The term $\frac{1}{h_{n-1}}$ is the $(n-1)$-th term of the $A$.$P$. of reciprocals,which is $\frac{1}{A} + (n-1)d' = \frac{1}{50} - \frac{n-1}{100(n+1)}$.
$\frac{1}{h_{n-1}} = \frac{2(n+1) - (n-1)}{100(n+1)} = \frac{2n+2-n+1}{100(n+1)} = \frac{n+3}{100(n+1)}$.
Thus,$h_{n-1} = \frac{100(n+1)}{n+3}$.
Finally,$a_2 h_{n-1} = \frac{50(n+3)}{n+1} \times \frac{100(n+1)}{n+3} = 5000$.
104
DifficultMCQ
Consider two positive numbers $a$ and $b$. If the arithmetic mean of $a$ and $b$ exceeds their geometric mean by $\frac{3}{2}$ and the geometric mean of $a$ and $b$ exceeds their harmonic mean by $\frac{6}{5}$,then the absolute value of $(a^2 - b^2)$ is equal to
A
$153$
B
$135$
C
$154$
D
$136$

Solution

(B) Let $A, G, H$ be the arithmetic,geometric,and harmonic means of $a$ and $b$ respectively.
Given $A = G + \frac{3}{2}$ and $G = H + \frac{6}{5}$.
We know that $G^2 = AH$,so $H = \frac{G^2}{A}$.
Substituting $H$ in the second equation: $G = \frac{G^2}{A} + \frac{6}{5} \implies G - \frac{6}{5} = \frac{G^2}{G + 3/2}$.
$(G - 1.2)(G + 1.5) = G^2 \implies G^2 + 1.5G - 1.2G - 1.8 = G^2$.
$0.3G = 1.8 \implies G = 6$.
Then $A = 6 + 1.5 = 7.5 = \frac{15}{2}$ and $H = 6 - 1.2 = 4.8 = \frac{24}{5}$.
Since $A = \frac{a+b}{2} = \frac{15}{2}$,we have $a+b = 15$.
Since $G = \sqrt{ab} = 6$,we have $ab = 36$.
The numbers $a$ and $b$ are roots of $x^2 - 15x + 36 = 0$.
$(x - 12)(x - 3) = 0$,so ${a, b} = {12, 3}$.
$|a^2 - b^2| = |144 - 9| = 135$.
105
DifficultMCQ
If $a, b, c$ are in $A.P.$ and $a^2, b^2, c^2$ are in $G.P.$ such that $a < b < c$ and $a+b+c = \frac{3}{4}$,then the value of $a$ is
A
$\frac{1}{4} - \frac{1}{3\sqrt{2}}$
B
$\frac{1}{4} - \frac{1}{4\sqrt{2}}$
C
$\frac{1}{4} - \frac{1}{\sqrt{2}}$
D
$\frac{1}{4} - \frac{1}{2\sqrt{2}}$

Solution

(D) Given $a, b, c$ are in $A.P.$,so $a+c = 2b$.
Given $a+b+c = \frac{3}{4}$,substituting $a+c=2b$ gives $3b = \frac{3}{4}$,so $b = \frac{1}{4}$.
Given $a^2, b^2, c^2$ are in $G.P.$,so $(b^2)^2 = a^2 c^2$,which implies $ac = \pm b^2 = \pm \frac{1}{16}$.
Since $a < b < c$,$ac$ must be negative,so $ac = -\frac{1}{16}$.
We have $a+c = 2b = \frac{1}{2}$ and $ac = -\frac{1}{16}$.
The quadratic equation for $a$ and $c$ is $x^2 - (a+c)x + ac = 0$,which is $x^2 - \frac{1}{2}x - \frac{1}{16} = 0$.
Multiplying by $16$,we get $16x^2 - 8x - 1 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{8 \pm \sqrt{64 - 4(16)(-1)}}{32} = \frac{8 \pm \sqrt{128}}{32} = \frac{8 \pm 8\sqrt{2}}{32} = \frac{1}{4} \pm \frac{\sqrt{2}}{4} = \frac{1}{4} \pm \frac{1}{2\sqrt{2}}$.
Since $a < b$,we choose the smaller value: $a = \frac{1}{4} - \frac{1}{2\sqrt{2}}$.
106
DifficultMCQ
If the arithmetic mean of two numbers $a$ and $b$,where $a > b > 0$,is five times their geometric mean,then $\frac{a + b}{a - b}$ is equal to
A
$\frac{\sqrt{6}}{2}$
B
$\frac{3\sqrt{2}}{4}$
C
$\frac{7\sqrt{3}}{12}$
D
$\frac{5\sqrt{6}}{12}$

Solution

(D) Given that the arithmetic mean $(A.M.)$ is five times the geometric mean $(G.M.)$:
$\frac{a + b}{2} = 5\sqrt{ab}$
Divide both sides by $\sqrt{ab}$:
$\frac{a + b}{\sqrt{ab}} = 10$
Let $x = \sqrt{\frac{a}{b}}$. Then $\frac{a}{b} = x^2$. The equation becomes:
$\frac{x^2 + 1}{x} = 10 \implies x^2 - 10x + 1 = 0$
Solving for $x$ using the quadratic formula:
$x = \frac{10 \pm \sqrt{100 - 4}}{2} = 5 \pm \sqrt{24} = 5 \pm 2\sqrt{6}$
Since $a > b$,$x > 1$,so $x = 5 + 2\sqrt{6}$.
Then $\frac{a}{b} = (5 + 2\sqrt{6})^2 = 25 + 24 + 20\sqrt{6} = 49 + 20\sqrt{6}$.
Using Componendo and Dividendo on $\frac{a}{b} = \frac{49 + 20\sqrt{6}}{1}$:
$\frac{a + b}{a - b} = \frac{49 + 20\sqrt{6} + 1}{49 + 20\sqrt{6} - 1} = \frac{50 + 20\sqrt{6}}{48 + 20\sqrt{6}} = \frac{25 + 10\sqrt{6}}{24 + 10\sqrt{6}}$
Rationalizing the denominator:
$\frac{(25 + 10\sqrt{6})(24 - 10\sqrt{6})}{24^2 - (10\sqrt{6})^2} = \frac{600 - 250\sqrt{6} + 240\sqrt{6} - 600}{576 - 600} = \frac{-10\sqrt{6}}{-24} = \frac{5\sqrt{6}}{12}$
107
DifficultMCQ
Let $G$ be the geometric mean of two positive numbers $a$ and $b,$ and $M$ be the arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b}.$ If $\frac{1}{M} : G$ is $4:5,$ then $a:b$ can be
A
$1:4$
B
$1:2$
C
$2:3$
D
$3:4$

Solution

(A) $G = \sqrt{ab}$
$M = \frac{\frac{1}{a} + \frac{1}{b}}{2} = \frac{a + b}{2ab}$
Given that $\frac{1}{M} : G = 4 : 5$,we have $\frac{2ab}{(a + b)\sqrt{ab}} = \frac{4}{5}$
$\Rightarrow \frac{a + b}{2\sqrt{ab}} = \frac{5}{4}$
Using Componendo and Dividendo:
$\frac{a + b + 2\sqrt{ab}}{a + b - 2\sqrt{ab}} = \frac{5 + 4}{5 - 4}$
$\Rightarrow \frac{(\sqrt{a} + \sqrt{b})^2}{(\sqrt{a} - \sqrt{b})^2} = 9$
$\Rightarrow \frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = 3$ (assuming $a > b$ for positive ratio)
$\Rightarrow \sqrt{a} + \sqrt{b} = 3\sqrt{a} - 3\sqrt{b}$
$\Rightarrow 4\sqrt{b} = 2\sqrt{a}$ $\Rightarrow \sqrt{\frac{a}{b}} = 2$ $\Rightarrow \frac{a}{b} = 4$
Since the ratio $a:b$ can be $1:4$ or $4:1$,and $1:4$ is given in the options,the correct answer is $1:4$.
108
DifficultMCQ
Given a sequence of $4$ numbers,the first three of which are in $G.P.$ and the last three are in $A.P.$ with a common difference of $6$. If the first and last terms in this sequence are equal,then the last term is:
A
$16$
B
$8$
C
$4$
D
$2$

Solution

(B) Let the sequence be $a, b, c, d$.
Since $a, b, c$ are in $G.P.$,we have $b^2 = ac$.
Since $b, c, d$ are in $A.P.$ with common difference $6$,we have $c - b = 6$ and $d - c = 6$.
Thus,$c = b + 6$ and $d = c + 6 = b + 12$.
Given that the first and last terms are equal,$a = d$,so $a = b + 12$,which implies $b = a - 12$.
Substituting $b = a - 12$ into $c = b + 6$,we get $c = (a - 12) + 6 = a - 6$.
Now,substitute $b$ and $c$ into the $G.P.$ condition $b^2 = ac$:
$(a - 12)^2 = a(a - 6)$
$a^2 - 24a + 144 = a^2 - 6a$
$144 = 18a$
$a = 8$.
Since $d = a$,the last term is $8$.
109
DifficultMCQ
Let $a, b$ and $c$ be the $7^{th}, 11^{th}$ and $13^{th}$ terms respectively of a non-constant $A.P.$ If these are also the three consecutive terms of a $G.P.$,then $\frac{a}{c}$ is equal to
A
$\frac{1}{2}$
B
$4$
C
$2$
D
$\frac{7}{13}$

Solution

(B) Let the first term of the $A.P.$ be $A$ and the common difference be $d$. Since the $A.P.$ is non-constant,$d \neq 0$.
Given $a = A + 6d$,$b = A + 10d$,and $c = A + 12d$.
Since $a, b, c$ are in $G.P.$,we have $b^2 = ac$.
Substituting the values: $(A + 10d)^2 = (A + 6d)(A + 12d)$.
$A^2 + 20Ad + 100d^2 = A^2 + 18Ad + 72d^2$.
$2Ad = -28d^2$.
Since $d \neq 0$,we divide by $2d$ to get $A = -14d$,or $\frac{A}{d} = -14$.
Now,$\frac{a}{c} = \frac{A + 6d}{A + 12d} = \frac{\frac{A}{d} + 6}{\frac{A}{d} + 12}$.
Substituting $\frac{A}{d} = -14$: $\frac{-14 + 6}{-14 + 12} = \frac{-8}{-2} = 4$.
110
DifficultMCQ
If three distinct numbers $a, b, c$ are in $G.P.$ and the equations $ax^2 + 2bx + c = 0$ and $dx^2 + 2ex + f = 0$ have a common root,then which one of the following statements is correct?
A
$\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in $A.P.$
B
$d, e, f$ are in $A.P.$
C
$\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in $G.P.$
D
$d, e, f$ are in $G.P.$

Solution

(A) Since $a, b, c$ are in $G.P.$,we have $b^2 = ac$.
The equation $ax^2 + 2bx + c = 0$ can be written as $ax^2 + 2\sqrt{ac}x + c = 0$,which is $(\sqrt{a}x + \sqrt{c})^2 = 0$.
Thus,the root of this equation is $x = -\frac{\sqrt{c}}{\sqrt{a}} = -\frac{b}{a}$.
Since this is a common root for $dx^2 + 2ex + f = 0$,we substitute $x = -\frac{b}{a}$ into the second equation:
$d(-\frac{b}{a})^2 + 2e(-\frac{b}{a}) + f = 0$
$d(\frac{b^2}{a^2}) - \frac{2eb}{a} + f = 0$
Multiplying by $a^2$,we get $db^2 - 2eab + fa^2 = 0$.
Substituting $b^2 = ac$,we get $dac - 2eab + fa^2 = 0$.
Dividing the entire equation by $ac$,we get $\frac{d}{a} - \frac{2e}{b} + \frac{f}{c} = 0$,which implies $\frac{d}{a} + \frac{f}{c} = 2(\frac{e}{b})$.
This condition indicates that $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in $A.P.$
111
MediumMCQ
If $A.M.$ and $G.M.$ of two positive numbers $a$ and $b$ are $10$ and $8$ respectively,find the numbers.
A
$4, 16$
B
$6, 14$
C
$2, 18$
D
$5, 15$

Solution

(A) Given that $A.M. = \frac{a+b}{2} = 10 \implies a+b = 20$ $(1)$
Given that $G.M. = \sqrt{ab} = 8 \implies ab = 64$ $(2)$
Using the identity $(a-b)^2 = (a+b)^2 - 4ab$,we get:
$(a-b)^2 = (20)^2 - 4(64) = 400 - 256 = 144$
Thus,$a-b = \pm 12$ $(3)$
Solving $(1)$ and $(3)$:
If $a-b = 12$,then $2a = 32 \implies a = 16$ and $b = 4$.
If $a-b = -12$,then $2a = 8 \implies a = 4$ and $b = 16$.
Therefore,the numbers are $4$ and $16$.
112
MediumMCQ
Find the value of $n$ so that $\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}$ may be the geometric mean between $a$ and $b$.
A
$n = -\frac{1}{2}$
B
$n = 0$
C
$n = 1$
D
$n = \frac{1}{2}$

Solution

(A) The geometric mean $(GM)$ of $a$ and $b$ is $\sqrt{ab}$.
By the given condition:
$\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}} = \sqrt{ab} = (ab)^{1/2}$
Squaring both sides,we obtain:
$\frac{(a^{n+1}+b^{n+1})^2}{(a^n+b^n)^2} = ab$
$a^{2n+2} + 2a^{n+1}b^{n+1} + b^{2n+2} = ab(a^{2n} + 2a^nb^n + b^{2n})$
$a^{2n+2} + 2a^{n+1}b^{n+1} + b^{2n+2} = a^{2n+1}b + 2a^{n+1}b^{n+1} + ab^{2n+1}$
Subtracting $2a^{n+1}b^{n+1}$ from both sides:
$a^{2n+2} + b^{2n+2} = a^{2n+1}b + ab^{2n+1}$
Rearranging the terms:
$a^{2n+2} - a^{2n+1}b = ab^{2n+1} - b^{2n+2}$
$a^{2n+1}(a - b) = b^{2n+1}(a - b)$
Assuming $a \neq b$,we divide by $(a - b)$:
$a^{2n+1} = b^{2n+1}$
$(\frac{a}{b})^{2n+1} = 1 = (\frac{a}{b})^0$
$2n + 1 = 0$
$n = -\frac{1}{2}$
113
Difficult
The sum of two numbers is $6$ times their geometric mean. Show that the numbers are in the ratio $(3+2 \sqrt{2}):(3-2 \sqrt{2})$.

Solution

(N/A) Let the two numbers be $a$ and $b$.
The geometric mean $(G.M.)$ is $\sqrt{ab}$.
According to the given condition:
$a+b = 6\sqrt{ab}$ --- $(1)$
Squaring both sides:
$(a+b)^2 = 36ab$
We know that $(a-b)^2 = (a+b)^2 - 4ab$.
Substituting the value from $(1)$:
$(a-b)^2 = 36ab - 4ab = 32ab$
$a-b = \sqrt{32}\sqrt{ab} = 4\sqrt{2}\sqrt{ab}$ --- $(2)$
Adding $(1)$ and $(2)$:
$2a = (6+4\sqrt{2})\sqrt{ab}$
$a = (3+2\sqrt{2})\sqrt{ab}$
Substituting $a$ into $(1)$:
$b = 6\sqrt{ab} - (3+2\sqrt{2})\sqrt{ab}$
$b = (3-2\sqrt{2})\sqrt{ab}$
Therefore,the ratio is:
$\frac{a}{b} = \frac{(3+2\sqrt{2})\sqrt{ab}}{(3-2\sqrt{2})\sqrt{ab}} = \frac{3+2\sqrt{2}}{3-2\sqrt{2}}$
Thus,the numbers are in the ratio $(3+2\sqrt{2}):(3-2\sqrt{2})$.
114
Difficult
If $A$ and $G$ are the $A.M.$ and $G.M.$ respectively between two positive numbers,prove that the numbers are $A \pm \sqrt{(A + G)(A - G)}$.

Solution

It is given that $A$ and $G$ are the $A.M.$ and $G.M.$ between two positive numbers.
Let these two positive numbers be $a$ and $b$.
$\therefore A.M. = A = \frac{a+b}{2}$ .........$(1)$
$G.M. = G = \sqrt{ab}$ .........$(2)$
From $(1)$ and $(2)$,we obtain:
$a+b = 2A$ .........$(3)$
$ab = G^2$ .........$(4)$
We know the identity $(a-b)^2 = (a+b)^2 - 4ab$.
Substituting the values from $(3)$ and $(4)$:
$(a-b)^2 = (2A)^2 - 4G^2 = 4A^2 - 4G^2 = 4(A^2 - G^2)$
$(a-b)^2 = 4(A+G)(A-G)$
$(a-b) = 2\sqrt{(A+G)(A-G)}$ .........$(5)$
Adding $(3)$ and $(5)$:
$2a = 2A + 2\sqrt{(A+G)(A-G)} \Rightarrow a = A + \sqrt{(A+G)(A-G)}$
Subtracting $(5)$ from $(3)$:
$2b = 2A - 2\sqrt{(A+G)(A-G)} \Rightarrow b = A - \sqrt{(A+G)(A-G)}$
Thus,the two numbers are $A \pm \sqrt{(A+G)(A-G)}$.
115
Difficult
If $p^{\text{th}}, q^{\text{th}}, r^{\text{th}},$ and $s^{\text{th}}$ terms of an $A.P.$ are in $G.P.,$ then show that $(p-q), (q-r),$ and $(r-s)$ are also in $G.P.$

Solution

(N/A) Let the $A.P.$ be with first term $a$ and common difference $d$.
The terms are $a_p = a + (p-1)d, a_q = a + (q-1)d, a_r = a + (r-1)d, a_s = a + (s-1)d$.
Since $a_p, a_q, a_r, a_s$ are in $G.P.,$ the common ratio is $k = \frac{a_q}{a_p} = \frac{a_r}{a_q} = \frac{a_s}{a_r}$.
Using the property of ratios,$\frac{a_q - a_r}{a_p - a_q} = \frac{a_q(1 - k)}{a_p(1 - k)} = \frac{a_q}{a_p} = k$.
Also,$a_q - a_r = (a + (q-1)d) - (a + (r-1)d) = (q-r)d$ and $a_p - a_q = (p-q)d$.
Thus,$k = \frac{(q-r)d}{(p-q)d} = \frac{q-r}{p-q}$.
Similarly,$\frac{a_r - a_s}{a_q - a_r} = \frac{a_r(1 - k)}{a_q(1 - k)} = \frac{a_r}{a_q} = k$.
Thus,$k = \frac{(r-s)d}{(q-r)d} = \frac{r-s}{q-r}$.
Since both ratios equal $k,$ we have $\frac{q-r}{p-q} = \frac{r-s}{q-r},$ which implies $(p-q), (q-r), (r-s)$ are in $G.P.$
116
Medium
If $a, b, c$ are in $G.P.$ and $a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{z}} = k,$ prove that $x, y, z$ are in $A.P.$

Solution

(N/A) Let $a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{z}} = k.$
Then $a = k^{x}, b = k^{y},$ and $c = k^{z}$ $(1).$
Since $a, b, c$ are in $G.P.,$ we have $b^{2} = ac$ $(2).$
Substituting $(1)$ into $(2),$ we get $(k^{y})^{2} = k^{x} \cdot k^{z}.$
This simplifies to $k^{2y} = k^{x+z}.$
Equating the exponents,we get $2y = x + z.$
Since $2y = x + z,$ it follows that $x, y, z$ are in $A.P.$
117
DifficultMCQ
The sum of three numbers in $G.P.$ is $56$. If we subtract $1, 7, 21$ from these numbers in that order,we obtain an arithmetic progression. Find the numbers.
A
$8, 16, 32$
B
$4, 16, 36$
C
$32, 16, 8$
D
$2, 16, 38$

Solution

(A) Let the three numbers in $G.P.$ be $a, ar, ar^2$.
From the given condition,$a + ar + ar^2 = 56 \Rightarrow a(1 + r + r^2) = 56$ ........$(1)$
Subtracting $1, 7, 21$ from these numbers,we get $a-1, ar-7, ar^2-21$,which are in $A.P.$
Therefore,$(ar-7) - (a-1) = (ar^2-21) - (ar-7)$
$ar - a - 6 = ar^2 - ar - 14$
$ar^2 - 2ar + a = 8$ $\Rightarrow a(r^2 - 2r + 1) = 8$ $\Rightarrow a(r-1)^2 = 8$ ........$(2)$
Dividing $(1)$ by $(2)$:
$\frac{a(1+r+r^2)}{a(r-1)^2} = \frac{56}{8} = 7$
$1 + r + r^2 = 7(r^2 - 2r + 1)$
$1 + r + r^2 = 7r^2 - 14r + 7$
$6r^2 - 15r + 6 = 0 \Rightarrow 2r^2 - 5r + 2 = 0$
$(2r-1)(r-2) = 0$
If $r=2$,$a(2-1)^2 = 8 \Rightarrow a=8$. The numbers are $8, 16, 32$.
If $r=1/2$,$a(1/2-1)^2 = 8$ $\Rightarrow a(1/4) = 8$ $\Rightarrow a=32$. The numbers are $32, 16, 8$.
118
Difficult
If $a$ and $b$ are the roots of $x^{2}-3x+p=0$ and $c$ and $d$ are roots of $x^{2}-12x+q=0$,where $a, b, c, d$ form a $G.P.$,prove that $(q+p):(q-p)=17:15$.

Solution

Given that $a$ and $b$ are the roots of $x^{2}-3x+p=0$,we have $a+b=3$ and $ab=p$ $(1)$.
Given that $c$ and $d$ are the roots of $x^{2}-12x+q=0$,we have $c+d=12$ and $cd=q$ $(2)$.
Since $a, b, c, d$ are in $G.P.$,let $a=x, b=xr, c=xr^{2}, d=xr^{3}$.
From $(1)$,$x+xr=3 \Rightarrow x(1+r)=3$.
From $(2)$,$xr^{2}+xr^{3}=12 \Rightarrow xr^{2}(1+r)=12$.
Dividing the two equations: $\frac{xr^{2}(1+r)}{x(1+r)} = \frac{12}{3}$ $\Rightarrow r^{2}=4$ $\Rightarrow r=\pm 2$.
Case $I$: If $r=2$,then $x(1+2)=3 \Rightarrow x=1$. Thus $a=1, b=2, c=4, d=8$. Then $p=ab=2$ and $q=cd=32$.
$\frac{q+p}{q-p} = \frac{32+2}{32-2} = \frac{34}{30} = \frac{17}{15}$.
Case $II$: If $r=-2$,then $x(1-2)=3 \Rightarrow x=-3$. Thus $a=-3, b=6, c=-12, d=24$. Then $p=ab=-18$ and $q=cd=-288$.
$\frac{q+p}{q-p} = \frac{-288-18}{-288+18} = \frac{-306}{-270} = \frac{17}{15}$.
Thus,in both cases,$(q+p):(q-p)=17:15$.
119
Difficult
The ratio of the $A.M.$ and $G.M.$ of two positive numbers $a$ and $b$ is $m: n.$ Show that $a: b = (m + \sqrt{m^{2} - n^{2}}) : (m - \sqrt{m^{2} - n^{2}}).$

Solution

Let the two numbers be $a$ and $b.$
$A.M. = \frac{a+b}{2}$ and $G.M. = \sqrt{ab}.$
According to the given condition,
$\frac{a+b}{2\sqrt{ab}} = \frac{m}{n}$
$\Rightarrow \frac{(a+b)^{2}}{4ab} = \frac{m^{2}}{n^{2}}$
$\Rightarrow (a+b)^{2} = \frac{4abm^{2}}{n^{2}}$
$\Rightarrow a+b = \frac{2\sqrt{ab}m}{n} \quad \dots(1)$
Using the identity $(a-b)^{2} = (a+b)^{2} - 4ab,$
$(a-b)^{2} = \frac{4abm^{2}}{n^{2}} - 4ab = \frac{4ab(m^{2}-n^{2})}{n^{2}}$
$\Rightarrow a-b = \frac{2\sqrt{ab}\sqrt{m^{2}-n^{2}}}{n} \quad \dots(2)$
Adding $(1)$ and $(2),$ we obtain $2a = \frac{2\sqrt{ab}}{n}(m + \sqrt{m^{2}-n^{2}}),$
$\Rightarrow a = \frac{\sqrt{ab}}{n}(m + \sqrt{m^{2}-n^{2}}).$
Substituting $a$ in $(1),$ we get $b = \frac{\sqrt{ab}}{n}(m - \sqrt{m^{2}-n^{2}}).$
Therefore,$\frac{a}{b} = \frac{m + \sqrt{m^{2}-n^{2}}}{m - \sqrt{m^{2}-n^{2}}}.$
Thus,$a:b = (m + \sqrt{m^{2}-n^{2}}) : (m - \sqrt{m^{2}-n^{2}}).$
120
Difficult
If $a, b, c$ are in $A.P.;$ $b, c, d$ are in $G.P.$ and $\frac{1}{c}, \frac{1}{d}, \frac{1}{e}$ are in $A.P.,$ prove that $a, c, e$ are in $G.P.$

Solution

(N/A) It is given that $a, b, c$ are in $A.P.$
$\therefore 2b = a + c$ .......$(1)$
It is given that $b, c, d$ are in $G.P.$
$\therefore c^{2} = bd$ .......$(2)$
Also,$\frac{1}{c}, \frac{1}{d}, \frac{1}{e}$ are in $A.P.$
$\therefore \frac{2}{d} = \frac{1}{c} + \frac{1}{e}$ .......$(3)$
We need to prove that $a, c, e$ are in $G.P.,$ i.e.,$c^{2} = ae.$
From $(1),$ $b = \frac{a+c}{2}.$
From $(2),$ $d = \frac{c^{2}}{b}.$
Substituting these values into $(3):$
$\frac{2}{\frac{c^{2}}{b}} = \frac{1}{c} + \frac{1}{e}$
$\frac{2b}{c^{2}} = \frac{e+c}{ce}$
$\frac{2(\frac{a+c}{2})}{c^{2}} = \frac{e+c}{ce}$
$\frac{a+c}{c^{2}} = \frac{e+c}{ce}$
$\frac{a+c}{c} = \frac{e+c}{e}$
$(a+c)e = c(e+c)$
$ae + ce = ce + c^{2}$
$c^{2} = ae$
Thus,$a, c, e$ are in $G.P.$
121
DifficultMCQ
Let $\frac{1}{16}, a$ and $b$ be in $G.P.$ and $\frac{1}{a}, \frac{1}{b}, 6$ be in $A.P.,$ where $a, b > 0$. Then $72(a + b)$ is equal to ...... .
A
$12$
B
$18$
C
$14$
D
$21$

Solution

(C) Since $\frac{1}{16}, a, b$ are in $G.P.$,we have $a^2 = \frac{b}{16}$,which implies $b = 16a^2$.
Since $\frac{1}{a}, \frac{1}{b}, 6$ are in $A.P.$,we have $\frac{2}{b} = \frac{1}{a} + 6$.
Substituting $b = 16a^2$ into the $A.P.$ equation:
$\frac{2}{16a^2} = \frac{1}{a} + 6$
$\frac{1}{8a^2} = \frac{1}{a} + 6$
Multiply by $8a^2$:
$1 = 8a + 48a^2$
$48a^2 + 8a - 1 = 0$
$(12a - 1)(4a + 1) = 0$
Since $a > 0$,we have $a = \frac{1}{12}$.
Then $b = 16 \times (\frac{1}{12})^2 = 16 \times \frac{1}{144} = \frac{1}{9}$.
Finally,$72(a + b) = 72(\frac{1}{12} + \frac{1}{9}) = 72(\frac{3 + 4}{36}) = 72(\frac{7}{36}) = 2 \times 7 = 14$.
122
MediumMCQ
If $x = \sum_{n=0}^{\infty} a^{n}$,$y = \sum_{n=0}^{\infty} b^{n}$,$z = \sum_{n=0}^{\infty} c^{n}$,where $a, b, c$ are in $A.P.$ and $|a| < 1, |b| < 1, |c| < 1$,$abc \neq 0$,then:
A
$x, y, z$ are in $A.P.$
B
$\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in $A.P.$
C
$x, y, z$ are in $G.P.$
D
$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1 - (a + b + c)$

Solution

(B) Given $x = \sum_{n=0}^{\infty} a^{n} = \frac{1}{1-a}$,$y = \frac{1}{1-b}$,and $z = \frac{1}{1-c}$.
From these,we get $a = 1 - \frac{1}{x}$,$b = 1 - \frac{1}{y}$,and $c = 1 - \frac{1}{z}$.
Since $a, b, c$ are in $A.P.$,we have $2b = a + c$.
Substituting the expressions in terms of $x, y, z$:
$2(1 - \frac{1}{y}) = (1 - \frac{1}{x}) + (1 - \frac{1}{z})$
$2 - \frac{2}{y} = 2 - (\frac{1}{x} + \frac{1}{z})$
$\frac{2}{y} = \frac{1}{x} + \frac{1}{z}$.
This condition implies that $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in $A.P.$
123
AdvancedMCQ
The arithmetic mean and the geometric mean of two distinct $2$-digit numbers $x$ and $y$ are two integers,one of which can be obtained by reversing the digits of the other (in base $10$ representation). Then,$x+y$ equals
A
$82$
B
$116$
C
$130$
D
$148$

Solution

(C) Let the two integers be $A = 10a+b$ and $G = 10b+a$,where $A$ is the arithmetic mean and $G$ is the geometric mean.
Given,$\frac{x+y}{2} = 10a+b$ and $\sqrt{xy} = 10b+a$.
Thus,$x+y = 2(10a+b)$ and $xy = (10b+a)^2$.
We know that $(x-y)^2 = (x+y)^2 - 4xy$.
Substituting the values,$(x-y)^2 = 4(10a+b)^2 - 4(10b+a)^2$.
$(x-y)^2 = 4[(10a+b)^2 - (10b+a)^2] = 4(10a+b-10b-a)(10a+b+10b+a)$.
$(x-y)^2 = 4(9a-9b)(11a+11b) = 4 \times 9 \times 11(a-b)(a+b) = 396(a-b)(a+b)$.
For $(x-y)^2$ to be a perfect square,$(a-b)(a+b)$ must be of the form $11 \times k^2$.
Since $a$ and $b$ are digits,$a+b \leq 18$ and $a-b < 10$. The only possibility is $a+b=11$ and $a-b=1$.
Solving $a+b=11$ and $a-b=1$ gives $2a=12 \Rightarrow a=6$ and $b=5$.
Thus,$x+y = 2(10(6)+5) = 2(65) = 130$.
124
DifficultMCQ
For two positive numbers $a$ and $b$,if $a, b$ and $\frac{1}{18}$ are in a geometric progression,while $\frac{1}{a}, 10$ and $\frac{1}{b}$ are in an arithmetic progression,then $16a + 12b$ is equal to $.........$.
A
$3$
B
$2$
C
$1$
D
$0$

Solution

(A) Given $a, b, \frac{1}{18}$ are in $GP$,so $b^2 = a \times \frac{1}{18} \implies a = 18b^2$ $(i)$.
Given $\frac{1}{a}, 10, \frac{1}{b}$ are in $AP$,so $\frac{1}{a} + \frac{1}{b} = 2 \times 10 = 20$.
$\frac{a+b}{ab} = 20 \implies a+b = 20ab$ $(ii)$.
Substitute $(i)$ into $(ii)$:
$18b^2 + b = 20(18b^2)b = 360b^3$.
Since $b > 0$,divide by $b$: $18b + 1 = 360b^2 \implies 360b^2 - 18b - 1 = 0$.
Using the quadratic formula: $b = \frac{18 \pm \sqrt{324 - 4(360)(-1)}}{2(360)} = \frac{18 \pm \sqrt{324 + 1440}}{720} = \frac{18 \pm \sqrt{1764}}{720} = \frac{18 \pm 42}{720}$.
Since $b > 0$,$b = \frac{60}{720} = \frac{1}{12}$.
Then $a = 18 \times (\frac{1}{12})^2 = 18 \times \frac{1}{144} = \frac{1}{8}$.
Finally,$16a + 12b = 16(\frac{1}{8}) + 12(\frac{1}{12}) = 2 + 1 = 3$.
125
MediumMCQ
The $8^{\text{th}}$ common term of the series $S_1 = 3 + 7 + 11 + 15 + 19 + \dots$ and $S_2 = 1 + 6 + 11 + 16 + 21 + \dots$ is $.......$.
A
$150$
B
$151$
C
$152$
D
$153$

Solution

(B) The series $S_1$ is an arithmetic progression with first term $a_1 = 3$ and common difference $d_1 = 4$. Its general term is $T_n = 3 + (n-1)4 = 4n - 1$.
The series $S_2$ is an arithmetic progression with first term $a_2 = 1$ and common difference $d_2 = 5$. Its general term is $T_m = 1 + (m-1)5 = 5m - 4$.
For a common term,$4n - 1 = 5m - 4$,which implies $4n + 3 = 5m$.
The first common term is $11$ (for $n=3, m=3$).
The common difference of the new series formed by common terms is the least common multiple of $d_1$ and $d_2$,which is $\text{lcm}(4, 5) = 20$.
The $k^{\text{th}}$ common term is given by $T_k = 11 + (k-1)20$.
For $k=8$,$T_8 = 11 + (8-1) \times 20 = 11 + 7 \times 20 = 11 + 140 = 151$.
126
DifficultMCQ
Let $0 < z < y < x$ be three real numbers such that $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in an arithmetic progression and $x, \sqrt{2}y, z$ are in a geometric progression. If $xy + yz + zx = \frac{3}{\sqrt{2}} xyz$,then $3(x + y + z)^2$ is equal to $............$.
A
$150$
B
$140$
C
$130$
D
$120$

Solution

(A) Given that $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in an arithmetic progression,we have $\frac{2}{y} = \frac{1}{x} + \frac{1}{z}$.
Given that $x, \sqrt{2}y, z$ are in a geometric progression,we have $(\sqrt{2}y)^2 = xz$,which implies $2y^2 = xz$.
Substituting $xz = 2y^2$ into the first equation: $\frac{2}{y} = \frac{x+z}{xz} = \frac{x+z}{2y^2}$,which simplifies to $x+z = 4y$.
Given the equation $xy + yz + zx = \frac{3}{\sqrt{2}} xyz$,we can rewrite it as $y(x+z) + xz = \frac{3}{\sqrt{2}} xyz$.
Substituting $x+z = 4y$ and $xz = 2y^2$: $y(4y) + 2y^2 = \frac{3}{\sqrt{2}} y(2y^2)$.
$4y^2 + 2y^2 = \frac{3}{\sqrt{2}} (2y^3) \implies 6y^2 = 3\sqrt{2} y^3$.
Since $y > 0$,we divide by $3y^2$: $2 = \sqrt{2}y$,so $y = \sqrt{2}$.
Then $x+z = 4y = 4\sqrt{2}$,so $x+y+z = 5y = 5\sqrt{2}$.
Finally,$3(x+y+z)^2 = 3(5\sqrt{2})^2 = 3(25 \times 2) = 3(50) = 150$.
127
DifficultMCQ
Let the first three terms $2, p$ and $q$,with $q \neq 2$,of a $G.P.$ be respectively the $7^{\text{th}}$,$8^{\text{th}}$ and $13^{\text{th}}$ terms of an $A.P.$ If the $5^{\text{th}}$ term of the $G.P.$ is the $n^{\text{th}}$ term of the $A.P.$,then $n$ is equal to
A
$151$
B
$169$
C
$177$
D
$163$

Solution

(D) Let the $A.P.$ be $a, a+d, a+2d, \dots$ and the $G.P.$ be $2, p, q, \dots$
Given that $2, p, q$ are the $7^{\text{th}}, 8^{\text{th}}, 13^{\text{th}}$ terms of the $A.P.$:
$2 = a + 6d \quad \dots(i)$
$p = a + 7d \quad \dots(ii)$
$q = a + 12d \quad \dots(iii)$
Subtracting $(i)$ from $(ii)$,we get $p - 2 = d$.
Subtracting $(ii)$ from $(iii)$,we get $q - p = 5d$.
Since $q - p = 5(p - 2)$,we have $q = 6p - 10$.
Since $2, p, q$ are in $G.P.$,$p^2 = 2q$.
Substituting $q = 6p - 10$,we get $p^2 = 2(6p - 10) \implies p^2 - 12p + 20 = 0$.
Solving for $p$,$(p - 10)(p - 2) = 0$,so $p = 10$ or $p = 2$.
If $p = 2$,then $q = 2$,which contradicts $q \neq 2$. Thus,$p = 10$.
Then $d = p - 2 = 8$ and $a = 2 - 6(8) = -46$.
The $G.P.$ is $2, 10, 50, 250, 1250, \dots$
The $5^{\text{th}}$ term of the $G.P.$ is $2 \times 5^4 = 1250$.
Let this be the $n^{\text{th}}$ term of the $A.P.$: $1250 = a + (n - 1)d$.
$1250 = -46 + (n - 1)8 \implies 1296 = (n - 1)8 \implies n - 1 = 162 \implies n = 163$.
128
MediumMCQ
Let three real numbers $a, b, c$ be in arithmetic progression and $a+1, b, c+3$ be in geometric progression. If $a > 10$ and the arithmetic mean of $a, b$ and $c$ is $8$,then the cube of the geometric mean of $a, b$ and $c$ is
A
$120$
B
$312$
C
$316$
D
$128$

Solution

(A) Since $a, b, c$ are in arithmetic progression,we have $2b = a + c$.
Given the arithmetic mean of $a, b, c$ is $8$,we have $\frac{a+b+c}{3} = 8$,which implies $a+b+c = 24$.
Substituting $a+c = 2b$,we get $2b + b = 24$,so $3b = 24$,which gives $b = 8$.
Then $a+c = 16$,so $c = 16 - a$.
Since $a+1, b, c+3$ are in geometric progression,we have $b^2 = (a+1)(c+3)$.
Substituting $b=8$ and $c=16-a$,we get $8^2 = (a+1)(16-a+3)$,which simplifies to $64 = (a+1)(19-a)$.
$64 = 19a - a^2 + 19 - a$,which leads to $a^2 - 18a + 45 = 0$.
Factoring the quadratic equation,we get $(a-15)(a-3) = 0$.
Since $a > 10$,we must have $a = 15$.
Then $c = 16 - 15 = 1$.
The numbers are $a=15, b=8, c=1$.
The geometric mean of $a, b, c$ is $(abc)^{1/3}$.
The cube of the geometric mean is $( (abc)^{1/3} )^3 = abc = 15 \times 8 \times 1 = 120$.
129
AdvancedMCQ
Let $A_1, G_1, H_1$ denote the arithmetic,geometric,and harmonic means,respectively,of two distinct positive numbers $a$ and $b$. For $n \geq 2$,let $A_n, G_n, H_n$ be the arithmetic,geometric,and harmonic means of $A_{n-1}$ and $H_{n-1}$ respectively.
$1.$ Which one of the following statements is correct?
$(A)$ $G_1 > G_2 > G_3 > \ldots$
$(B)$ $G_1 < G_2 < G_3 < \ldots$
$(C)$ $G_1 = G_2 = G_3 = \ldots$
$(D)$ $G_1 < G_3 < G_5 < \ldots$ and $G_2 > G_4 > G_6 > \ldots$
$2.$ Which of the following statements is correct?
$(A)$ $A_1 > A_2 > A_3 > \ldots$
$(B)$ $A_1 < A_2 < A_3 < \ldots$
$(C)$ $A_1 > A_3 > A_5 > \ldots$ and $A_2 < A_4 < A_6 < \ldots$
$(D)$ $A_1 < A_3 < A_5 < \ldots$ and $A_2 > A_4 > A_6 > \ldots$
$3.$ Which of the following statements is correct?
$(A)$ $H_1 > H_2 > H_3 > \ldots$
$(B)$ $H_1 < H_2 < H_3 < \ldots$
$(C)$ $H_1 > H_3 > H_5 > \ldots$ and $H_2 < H_4 < H_6 < \ldots$
$(D)$ $H_1 < H_3 < H_5 < \ldots$ and $H_2 > H_4 > H_6 > \ldots$
Give the answers for questions $1, 2,$ and $3.$
A
$C, A, B$
B
$C, B, A$
C
$A, A, B$
D
$C, A, C$

Solution

(C, A, B) $1.$ Given $A_n = \frac{A_{n-1} H_{n-1}}{2}$,$G_n = \sqrt{A_{n-1} H_{n-1}}$,and $H_n = \frac{2 A_{n-1} H_{n-1}}{A_{n-1} H_{n-1}}$.
Note that $G_n^2 = A_{n-1} H_{n-1} = A_n H_n$. Also,$G_n = \sqrt{A_{n-1} H_{n-1}} = \sqrt{G_{n-1}^2} = G_{n-1}$.
Thus,$G_1 = G_2 = G_3 = \ldots = \sqrt{ab}$.
$2.$ Since $A_n$ is the arithmetic mean of $A_{n-1}$ and $H_{n-1}$,and $A_{n-1} > H_{n-1}$ for $n \geq 2$,we have $A_{n-1} > A_n > H_{n-1}$.
Since $H_{n-1} < H_n < A_n < A_{n-1}$,it follows that $A_1 > A_2 > A_3 > \ldots$.
$3.$ Since $H_n$ is the harmonic mean of $A_{n-1}$ and $H_{n-1}$,and $A_{n-1} > H_{n-1}$,we have $A_{n-1} > H_n > H_{n-1}$.
Since $H_n > H_{n-1}$,it follows that $H_1 < H_2 < H_3 < \ldots$.
130
AdvancedMCQ
Let $b_i > 1$ for $i = 1, 2, \ldots, 101$. Suppose $\log _e b_1, \log _e b_2, \ldots, \log _e b_{101}$ are in Arithmetic Progression $(A.P.)$ with the common difference $\log _e 2$. Suppose $a_1, a_2, \ldots, a_{101}$ are in $A.P.$ such that $a_1 = b_1$ and $a_{51} = b_{51}$. If $t = b_1 + b_2 + \cdots + b_{51}$ and $s = a_1 + a_2 + \cdots + a_{51}$,then:
A
$s > t$ and $a_{101} > b_{101}$
B
$s > t$ and $a_{101} < b_{101}$
C
$s < t$ and $a_{101} > b_{101}$
D
$s < t$ and $a_{101} < b_{101}$

Solution

(B) Given $\log _e b_1, \log _e b_2, \ldots, \log _e b_{101}$ are in $A.P.$,it follows that $b_1, b_2, \ldots, b_{101}$ are in $G.P.$ with common ratio $r = e^{\log _e 2} = 2$.
Let $a_1 = b_1 = a$. Since $a_{51} = b_{51}$,we have $a + 50d = a \cdot r^{50} = a \cdot 2^{50}$,so $d = \frac{a(2^{50} - 1)}{50}$.
The sum $t$ of the first $51$ terms of the $G.P.$ is $t = b_1 \frac{r^{51} - 1}{r - 1} = a(2^{51} - 1)$.
The sum $s$ of the first $51$ terms of the $A.P.$ is $s = \frac{51}{2}(2a + 50d) = \frac{51}{2}(2a + a(2^{50} - 1)) = \frac{51}{2}a(2^{50} + 1)$.
Comparing $s$ and $t$: $s = \frac{51}{2}a(2^{50} + 1) \approx 25.5 \cdot a \cdot 2^{50}$ and $t = a(2^{51} - 1) \approx 2 \cdot a \cdot 2^{50}$. Thus,$s > t$.
For the $101^{st}$ terms: $a_{101} = a + 100d = a + 2a(2^{50} - 1) = a(2^{51} - 1)$.
$b_{101} = b_1 \cdot r^{100} = a \cdot 2^{100}$.
Since $2^{100} > 2^{51} - 1$,we have $b_{101} > a_{101}$.
131
EasyMCQ
If $a, b, c$ are in $AP$,and $b-a, c-b, a$ are in $GP$,then $a: b: c$ is
A
$1: 2: 3$
B
$1: 3: 5$
C
$2: 3: 4$
D
$1: 2: 4$

Solution

(A) Given that $a, b, c$ are in $AP$,so $2b = a + c$.
Since $b-a, c-b, a$ are in $GP$,we have $(c-b)^2 = (b-a)a$.
In an $AP$,$b-a = c-b = d$ (common difference).
Substituting $c-b = b-a$ into the $GP$ condition:
$(b-a)^2 = (b-a)a$.
Assuming $b \neq a$,we get $b-a = a$,which implies $b = 2a$.
Substituting $b = 2a$ into $2b = a + c$,we get $2(2a) = a + c$,so $4a = a + c$,which implies $c = 3a$.
Thus,$a: b: c = a: 2a: 3a = 1: 2: 3$.
132
EasyMCQ
Let $x_{1}, x_{2}$ be the roots of $x^{2}-3x+a=0$ and $x_{3}, x_{4}$ be the roots of $x^{2}-12x+b=0$. If $x_{1} < x_{2} < x_{3} < x_{4}$ and $x_{1}, x_{2}, x_{3}, x_{4}$ are in $GP$,then $ab$ equals:
A
$24/5$
B
$64$
C
$16$
D
$8$

Solution

(B) Given that $x_{1}, x_{2}$ are roots of $x^{2}-3x+a=0$,so $x_{1}+x_{2}=3$ and $x_{1}x_{2}=a$.
Given that $x_{3}, x_{4}$ are roots of $x^{2}-12x+b=0$,so $x_{3}+x_{4}=12$ and $x_{3}x_{4}=b$.
Since $x_{1}, x_{2}, x_{3}, x_{4}$ are in $GP$,let them be $A, AR, AR^{2}, AR^{3}$.
Then $x_{1}+x_{2} = A(1+R) = 3$ and $x_{3}+x_{4} = AR^{2}(1+R) = 12$.
Dividing the two equations,we get $R^{2} = 12/3 = 4$,so $R = 2$ (since $x_{1} < x_{2} < x_{3} < x_{4}$ implies $R > 0$).
Substituting $R=2$ into $A(1+R)=3$,we get $A(3)=3$,so $A=1$.
The terms are $1, 2, 4, 8$.
Thus,$a = x_{1}x_{2} = 1 \times 2 = 2$ and $b = x_{3}x_{4} = 4 \times 8 = 32$.
Therefore,$ab = 2 \times 32 = 64$.
133
EasyMCQ
Given an $A.P.$ and a $G.P.$ with positive terms,where the first and second terms of both progressions are equal. If $a_n$ and $b_n$ are the $n^{\text{th}}$ terms of the $A.P.$ and $G.P.$ respectively,then:
A
$a_n > b_n$ for all $n > 2$
B
$a_n < b_n$ for all $n > 2$
C
$a_n = b_n$ for some $n > 2$
D
$a_n = b_n$ for some odd $n$

Solution

(B) Let the first term be $a_1$ and the second term be $a_2$. Since terms are positive,$a_1 > 0$ and $a_2 > 0$.
For the $A.P.$,the common difference $d = a_2 - a_1$. Thus,$a_n = a_1 + (n - 1)(a_2 - a_1)$.
For the $G.P.$,the common ratio $r = \frac{a_2}{a_1}$. Thus,$b_n = a_1 \left(\frac{a_2}{a_1}\right)^{n-1}$.
For $n = 3$,$a_3 = a_1 + 2(a_2 - a_1) = 2a_2 - a_1$.
For $n = 3$,$b_3 = a_1 \left(\frac{a_2}{a_1}\right)^2 = \frac{a_2^2}{a_1}$.
Consider $b_3 - a_3 = \frac{a_2^2}{a_1} - (2a_2 - a_1) = \frac{a_2^2 - 2a_1a_2 + a_1^2}{a_1} = \frac{(a_2 - a_1)^2}{a_1}$.
Since $a_1 > 0$ and $(a_2 - a_1)^2 \ge 0$,$b_3 - a_3 \ge 0$,implying $b_3 \ge a_3$. If $a_1 \neq a_2$,then $b_3 > a_3$. By induction or properties of convex functions,$b_n > a_n$ for all $n > 2$.
134
EasyMCQ
The Geometric Mean $(G.M.)$ and Harmonic Mean $(H.M.)$ of two numbers are $10$ and $8$ respectively. The numbers are:
A
$5, 20$
B
$4, 25$
C
$2, 50$
D
$1, 100$

Solution

(A) Let the two numbers be $a$ and $b$.
Given that the $G.M. = \sqrt{ab} = 10$,so $ab = 100$.
Given that the $H.M. = \frac{2ab}{a+b} = 8$.
Substituting $ab = 100$ into the $H.M.$ equation:
$\frac{2(100)}{a+b} = 8$ $\Rightarrow \frac{200}{a+b} = 8$ $\Rightarrow a+b = \frac{200}{8} = 25$.
Now we have $a+b = 25$ and $ab = 100$.
The quadratic equation with roots $a$ and $b$ is $x^2 - (a+b)x + ab = 0$.
$x^2 - 25x + 100 = 0$.
$(x-20)(x-5) = 0$.
Thus,the numbers are $5$ and $20$.
135
EasyMCQ
The value of $n$ for which $\frac{x^{n+1}+y^{n+1}}{x^{n}+y^{n}}$ is the geometric mean of $x$ and $y$ is
A
$n=-\frac{1}{2}$
B
$n=\frac{1}{2}$
C
$n=1$
D
$n=-1$

Solution

(A) Given that $\frac{x^{n+1}+y^{n+1}}{x^n+y^n} = \sqrt{xy}$.
This implies $x^{n+1} + y^{n+1} = (xy)^{1/2} (x^n + y^n)$.
$x^{n+1} + y^{n+1} = x^{n+1/2} y^{1/2} + x^{1/2} y^{n+1/2}$.
Rearranging the terms,we get $x^{n+1} - x^{n+1/2} y^{1/2} = x^{1/2} y^{n+1/2} - y^{n+1}$.
$x^{n+1/2} (x^{1/2} - y^{1/2}) = y^{n+1/2} (x^{1/2} - y^{1/2})$.
Assuming $x \neq y$,we can divide both sides by $(x^{1/2} - y^{1/2})$:
$x^{n+1/2} = y^{n+1/2}$.
$(x/y)^{n+1/2} = 1$.
Since $(x/y)^0 = 1$,we have $n + 1/2 = 0$.
Therefore,$n = -1/2$.
136
EasyMCQ
For what value of $m$ is $\frac{a^{m+1}+b^{m+1}}{a^{m}+b^{m}}$ the arithmetic mean of $a$ and $b$?
A
$1$
B
$0$
C
$2$
D
None

Solution

(B) The arithmetic mean of $a$ and $b$ is given by $\frac{a+b}{2}$.
Given that $\frac{a^{m+1}+b^{m+1}}{a^m+b^m} = \frac{a+b}{2}$.
Cross-multiplying the terms,we get:
$2(a^{m+1} + b^{m+1}) = (a+b)(a^m + b^m)$
$2a^{m+1} + 2b^{m+1} = a^{m+1} + ab^m + ba^m + b^{m+1}$
Subtracting $a^{m+1} + b^{m+1}$ from both sides:
$a^{m+1} + b^{m+1} = ab^m + ba^m$
$a^{m+1} - ba^m = ab^m - b^{m+1}$
$a^m(a - b) = b^m(a - b)$
If $a \neq b$,we can divide by $(a - b)$:
$a^m = b^m$
$\left(\frac{a}{b}\right)^m = 1$
Since $1 = (\frac{a}{b})^0$,we have $m = 0$.
137
MediumMCQ
If the first and $(2n-1)$-th terms of an $AP$,$GP$,and $HP$ are equal and their $n$-th terms are respectively $a, b, c$,then always
A
$a=b=c$
B
$a \geq b \geq c$
C
$a+c=b$
D
$ac-b^2=0$

Solution

(D) Let the first term be $x$ and the $(2n-1)$-th term be $y$.
Since $a, b, c$ are the $n$-th terms of $AP, GP, HP$ respectively,they are the Arithmetic Mean,Geometric Mean,and Harmonic Mean of $x$ and $y$.
We know that for any two positive numbers $x$ and $y$,the relationship between their means is $AM \geq GM \geq HM$.
Therefore,$a \geq b \geq c$.
Also,the relationship between $AM, GM,$ and $HM$ is $AM \cdot HM = GM^2$,which implies $a \cdot c = b^2$ or $ac - b^2 = 0$.
Since the question asks for the relationship that is 'always' true,both $a \geq b \geq c$ and $ac = b^2$ are correct. However,in standard competitive mathematics,$ac = b^2$ is the specific identity derived from the definition of these means.
138
DifficultMCQ
Suppose $a, b, c$ are in $A.P.$ and $a^{2}, 2b^{2}, c^{2}$ are in $G.P.$ If $a < b < c$ and $a+b+c=1,$ then $9(a^{2}+b^{2}+c^{2})$ is equal to . . . . . . .
A
$7$
B
$9$
C
$12$
D
$15$

Solution

(B) Given $a, b, c$ are in $A.P.,$ let $a = b - d$ and $c = b + d.$
Since $a+b+c=1,$ we have $(b-d) + b + (b+d) = 1,$ which implies $3b = 1$ or $b = \frac{1}{3}.$
Given $a^{2}, 2b^{2}, c^{2}$ are in $G.P.,$ we have $(2b^{2})^{2} = a^{2}c^{2},$ so $4b^{4} = (ac)^{2}.$
Substituting $a = b-d$ and $c = b+d,$ we get $4b^{4} = ((b-d)(b+d))^{2} = (b^{2}-d^{2})^{2}.$
Taking the square root,$b^{2}-d^{2} = \pm 2b^{2}.$
Case $1: b^{2}-d^{2} = 2b^{2} \Rightarrow d^{2} = -b^{2}$ (not possible for real $d$).
Case $2: b^{2}-d^{2} = -2b^{2} \Rightarrow d^{2} = 3b^{2} = 3(\frac{1}{9}) = \frac{1}{3}.$
Since $a < b < c,$ $d$ must be positive,so $d = \frac{1}{\sqrt{3}}.$
Now,$a^{2}+b^{2}+c^{2} = (b-d)^{2} + b^{2} + (b+d)^{2} = 3b^{2} + 2d^{2}.$
Substituting values: $3(\frac{1}{9}) + 2(\frac{1}{3}) = \frac{1}{3} + \frac{2}{3} = 1.$
Therefore,$9(a^{2}+b^{2}+c^{2}) = 9(1) = 9.$

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