$\therefore \quad \log x, \log y, \log z$ are in A.P.
If the $n^{\text {th }}$ term of the G.P. $5,10,20, \ldots$ is equal to the $n^{\text {th }}$ term of the G.P. $1280,640,320, \ldots$, find the value of $n$.
SOLUTION $\begin{aligned} 5,10,20, \ldots \text { is a G.P. }& \\\\ a &=5 \\\\ r &=\dfrac{10}{5}=2 . \\\\ u_{n} &=a r^{n-1} \\\\ &=5(2)^{n-1} \\\\ 1280,640,320, \ldots \text { is a G.P. } \\\\ a &=1280 \\\\ r &=\dfrac{640}{1280}=\dfrac{1}{2}\\\\ u_{n} &=a r^{n-1} \\\\ &=1280\left(\dfrac{1}{2}\right)^{n-1} \\\\ &=1280(2)^{1-n}\\\\ \text{By the problem,}&\\\\ 5(2)^{n-1} &=1280(2)^{1-n} \\\\ \therefore \dfrac{2^{n-1}}{2^{1-n}} &=256 \\\\ 2^{2 n-2} &=2^{8} \\\\ 2 n-2 &=8 \\\\ 2 n &=10 \\\\ n &=5 \end{aligned}$
Find three numbers in G.P. whose sum is 19 and whose product is $216 .$
SOLUTION Let the three numbers in GP be $a$, ar, $a r^{2}\\\\$.
By the problem,
$\begin{aligned} &\\ a+a r+a r^{2}&=19 \\\\ a\left(1+r+r^{2}\right)&=19\dots(1)\\\\ a\cdot ar\cdot a r^{2} &=216 \\\\ a^{3} r^{3} &=6^{3} \\\\ (a r)^{3} &=6^{3}\\\\ ar &=6\\\\ a&=\dfrac{6}{r}\dots(2)\\\\ \end{aligned}$
If in a GP the $(p+q)^{\text {th }}$ term is $a$ and the $(p-q)^{\text {th }}$ term is $b$, prove that the $p^{\text {th }}$ term is $\sqrt{a b}$.
Solution
Let the first term and the common ratio of given G.P. be $A$ and $R$ respectively.
By the problem,
$\begin{aligned} &\\ u_{p+q}&=a \\\\ A R^{p+q-1}&=a\ldots(1) \\\\ u_{p-q}&=b\\\\ A R^{p-q-1} &=b \ldots(2)\\\\ \text { Miltiply equation (1) } & \text { by equation (2), }\\\\ A R^{p+q-1} \cdot A R^{p-q-1} &=a b \\\\ A^{2} R^{2 p-2} &=a b \\\\ \left(A R^{p-1}\right)^{2} &=a b \\\\ A R^{p-1} &=\sqrt{a b} \\\\ \therefore\ u_p &=\sqrt{a b} \end{aligned}$
The third term of a G.P. is $6 \dfrac{1}{4}$ and the $7^{\text {th }}$ term is the reciprocal of the third. Which term of this GP is unity?
SOlUTION
$\begin{aligned} \text { In a G.P., }& \\\\ u_{3}&=6 \dfrac{1}{4} \\\\ a r^{2}&=\dfrac{25}{4}\ldots(1) \\\\ u_{7}&=\dfrac{4}{25} \\\\ a r^{6}&=\dfrac{4}{25}\ldots(2)\\\\ \therefore \dfrac{a r^{6}}{a r^{2}} &=\dfrac{\frac{4}{25}}{\frac{25}{4}} \\\\ r^{4} &=\dfrac{16}{625} \\\\ r &=\dfrac{2}{5} \\\\ \therefore\ a\left(\dfrac{2}{5}\right)^{2} &=\dfrac{25}{4} \\\\ \dfrac{4 a}{25} &=\dfrac{25}{4}\\\\ \therefore\ a &=\dfrac{625}{16} \\\\ \therefore\ a r^{4} &=\dfrac{625}{16} \times \dfrac{16}{625} \\\\ \therefore\ u_{5} &=1 \end{aligned}$
If $\dfrac{1}{x+y}, \dfrac{1}{2 y}, \dfrac{1}{y+z}$ are the three consecutive terms of an A.P., prove that $x, y, z$ are the three consecutive terms of a G.P.
SOLUTION
$\dfrac{1}{x+y},\ \dfrac{1}{2 y},\ \dfrac{1}{y+z}$ are in A.P.
If the arithmetic mean and geometric mean between the two numbers are $30$ and $18$ respectively, find the two numbers
SOLUTION Let the two numbers be $a$ and $b\\\\ $.
A.M between $a$ and $b=30\\\\ $
$\begin{aligned} \dfrac{a+b}{2} &=30 \\\\ a+b &=60 \\\\ b &=60-a\\\\ \end{aligned}$
G.M. between $a$ and $b=18\\\\ $
$\begin{aligned} \sqrt{a b}&=18\\\\ a b &=324 \\\\ a(60-a) &=324 \\\\ 60 a-a^{2} &=324 \\\\ a^{2}-60 a+32 y &=0 \\\\ (a-6)(a-54) &=0 \\\\ a=6 \text { or } a &=54\\\\ \end{aligned}$
When $a=6, b=60-6=54\\\\ $.
When $a=54, b=60-54=6\\\\ $.
$\therefore$ The two numbers are $6$ and $54$.
Let $a, b, c$ are three numbers between $2$ and $18$ , such that their sum is $25$. If $2, a, b$ are in A.P. and $b, c, 18$ are in G.P., then find $c$.
SOLUTION $2, a, b$ are in A.P.
$\begin{aligned} &\\ \therefore a &=\dfrac{2+b}{2} \\\\ &=1+\dfrac{b}{2}\\\\ \end{aligned}$
An A.P. has first term $a$ and common difference $d, d \neq 0$. If the $3^{\text {rd }}, 4^{\text {th }}$ and $7^{\text {th }}$ terms of this A.P. are the first three terms of a G.P., show that $a=-\dfrac{3}{2} d$. Hence show that the $4^{\text {th }}$ term of the G.P. is the $16^{\text {th }}$ term of the A.P.
SOlution
In an A.P.,
$\begin{aligned} &\\ a&=\text { first term } \\\\ d&=\text { common difference } \\\\ u_{3}&=a+2 d \\\\ u_{4}&=a+3 d \\\\ u_{7}&=a+6 d \\\\ \end{aligned}$
By the problem,
$a+2 d, a+3 d, a+6 d$ are in a G.P.
$\begin{aligned} &\\ \therefore \dfrac{a+3 d}{a+2 d} &=\dfrac{a+6 d}{a+3 d} \\\\ \therefore a^{2}+6 a d+9 d^{2} &=a^{2}+8 a d+12 d^{2} \\\\ \therefore \quad-2 a d &=3 d^{2} \\\\ a &=-\dfrac{3}{2} d\\\\ \end{aligned}$
Let the common ratio of G.P. be $r$.
$\begin{aligned} &\\ \text { then, } r &=\dfrac{a+3 d}{a+2 d} \\\\ &=\dfrac{-\dfrac{3}{2} d+3 d}{-\dfrac{3}{2} d+2 d} \\\\ &=3 \\\\ \therefore\ 4^{\text {th }} \text { term of G.P.} &=3(a+6 d) \\\\ &=3\left(-\dfrac{3}{2} d+6 d\right) \\\\ &=\dfrac{27}{2} d\\\\ 16^{\text {th }} \text { term of A.P. } &=a+15 d \\\\ &=-\dfrac{3}{2} d+15 d \\\\ &=\dfrac{27}{2} d\\\\ \end{aligned}$
Hence, proved.
In a set of four numbers, the first three are in G.P. and the last three are in A.P. with a common difference 6 . If the first number is the same as the $4^{\text {th }}$, find the four numbers.
SOLUTION
Let $A=\{w, x, y, z\}\\\\$
By the problem,
$w, x, y$ are in GP.
$\\ \dfrac{x}{w}=\dfrac{y}{x}\\\\$
$x, y, z$ are in A.P.
Let the common difference be $d.\\\\$
$\begin{aligned} \therefore\ x &=x, y=x+6, z=x+12 \\\\ w &=z(\text { given }) \\\\ \therefore\ w &=x+12 \\\\ \dfrac{x}{x+12} &=\dfrac{x+6}{x} \\\\ x^{2} &=x^{2}+18 x+72 \\\\ x &=-4 \\\\ \therefore\ y &=2, z=w=8 \end{aligned}$
If the A.M. between two positive numbers $a$ and $b$, where $a>b$, is twice the G.M. between them, prove that $a: b=\left(2+\sqrt{3}\right): \left(2-\sqrt{3}\right)$
SOLUTION A.M. between $a$ and $b=\dfrac{a+b}{2}\\\\ $
G.M. between $a$ and $b =\sqrt{a b} \\\\ $
$\begin{aligned} \text{By the problem} &\\\\ \text{A.M.} &= 2\cdot\text {G.M.}\\\\ \dfrac{a+b}{2}&=2 \sqrt{a b} \\\\ a+b&=4 \sqrt{a b} \\\\ (a+b)^{2}&=16 a b \dots(1)\\\\ a^{2}+2 a b+b^{2}&=16 a b \\\\ a^{2}-14 a b+b^{2}&=0\\\\ a^{2}-2 a b+b^{2}&=12 a b \\\\ (a-b)^{2}&=12 a b\ldots(2) \\\\ \text{By } (1)\div (2), &\\\\ \dfrac{(a+b)^{2}}{(a-b)^{2}}&=\dfrac{16 a b}{12 a b} \\\\ \left(\dfrac{a+b}{a-b}\right)^{2}&=\dfrac{4}{3}\\\\ \dfrac{a+b}{a-b}&=\dfrac{2}{\sqrt{3}} \\\\ 2 a-2 b&=\sqrt{3} a+\sqrt{3} b \\\\ 2 a-\sqrt{3} a&=2 b+\sqrt{3} b \\\\ a(2-\sqrt{3})&=b(2+\sqrt{3}) \\\\ \therefore \dfrac{a}{b}&=\dfrac{2+\sqrt{3}}{2-\sqrt{3}} \end{aligned}$
If $A_{1}, A_{2}$ are the two A.M.s between two numbers $a$ and $b$ and $G_{1}, G_{2}$ are two G.M.s between the same two numbers, then prove that $\dfrac{A_{1}+A_{2}}{G_{1} G_{2}}=\dfrac{a+b}{a b}$.
SOLUTION $A_{1}, A_{2}$ are the two A.M.s between two numbers $a$ and $b\\\\ $
If $A$ and $G$ be A.M and G.M respectively between two positive numbers. Prove that the numbers are $A \pm \sqrt{A^{2}-G^{2}}$
SOLUTION
$\begin{aligned} \therefore\ \dfrac{a+b}{2} &=A \\\\ a+b &=2 A \\\\ \sqrt{a b} &=G \\\\ a b &=G^{2} \\\\ \left(a+b^{2}\right) &=4 A^{2} \\\\ a^{2}+2 a b+b^{2} &=4 A^{2}\\\\ a^{2}+b^{2} &=4 A^{2}-2 a b \\\\ a^{2}+b^{2}-2 a b &=4 A^{2}-4 a b \\\\ (a-b)^{2} &=4 A^{2}-4 G^{2} \\\\ a-b &=2 \sqrt{A^{2}-G^{2}} \\\\ \text{By } (1)+(2), & \\\\ 2 a &=2 A+\sqrt{A^{2}-G^{2}} \\\\ \therefore\ a &=A+\sqrt{A^{2}-G^{2}} \\\\ \text{By } (1)-(2),\\\\ 2b &=2 A-2 \sqrt{A^{2}-G^{2}} \\\\ \therefore\ b &=A-\sqrt{A^{2}-G^{2}} \end{aligned}$
If the ratio of A.M. and G.M. between two positive numbers is $m: n$, then prove that the numbers are in the ratio $\left(m+\sqrt{m^{2}-n^{2}}\right):\left(m-\sqrt{m^{2}-n^{2}}\right)$.
SOLUTION
Let the two numbers be $a$ and $b\\\\ $.
A.M. between $a$ and $b=\dfrac{a+b}{2}\\\\ $
G.M. between $a$ and $b=\sqrt{a b}\\\\ $
By the problem,
$\begin{aligned} &\\ \dfrac{\text{A.M}}{\text{G.M}}&=\dfrac{m}{n} \\\\ \dfrac{a+b}{2 \sqrt{a b}}&=\dfrac{m}{n}\\\\ \text{Let } a+b=k m, 2 \sqrt{a b}&=k n\\\\ \therefore\ (a+b)^{2} &=k^{2} m^{2} \\\\ 4 a b &=k^{2} n^{2} \\\\ a^{2}+2 a b+b^{2} &=k^{2} m^{2} \\\\ a^{2}+2 a b+b^{2}-4 a b &= k^{2} n^{2}-4 a b \\\\ a^{2}-2 a b+b^{2} &=k^{2} m^{2}-k^{2} n^{2} \\\\ (a-b)^{2} &=k^{2}\left(n^{2}-n^{2}\right) \\\\ a-b &=k \sqrt{m^{2}-n^{2}}\\\\ \dfrac{a+b}{a-b} &=\dfrac{k m}{k \sqrt{m^{2}-n^{2}}} \\\\ &=\dfrac{m}{\sqrt{m^{2}-n^{2}}}\\\\ \end{aligned}$
If $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of a GP be $a, b, c(a, b, c>0)$, then prove that $(q-r) \log a+(r-p) \log b+(p-q) \log c=0$.
SOLUTION
Let the first term be $A$ and the cominon ratio be $R$ of the given G.P.
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